{"id":16572,"date":"2019-10-03T19:20:03","date_gmt":"2019-10-03T19:20:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/applications-of-systems-of-equations\/"},"modified":"2021-01-19T12:09:11","modified_gmt":"2021-01-19T12:09:11","slug":"applications-of-systems-of-equations","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/chapter\/applications-of-systems-of-equations\/","title":{"raw":"10.3.a - Mixture Problems","rendered":"10.3.a &#8211; Mixture Problems"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve mixture problems<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 id=\"title1\">Write a system of linear equations representing a mixture problem, solve the system and interpret the results<\/h2>\r\nOne application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. \u00a0A solution is a mixture of two or more different substances like water and salt or vinegar and oil. \u00a0Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. \u00a0There are many other disciplines that use solutions as well.\r\n\r\nThe concentration or strength of a liquid solution is often described\u00a0\u00a0as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have [latex]50[\/latex] grams of salt in a [latex]100mL[\/latex] of water you have a [latex]50\\%[\/latex] salt solution based on the following ratio:\r\n<p style=\"text-align: center\">[latex]\\frac{50\\text{ grams }}{100\\text{ mL }}=0.50\\frac{\\text{ grams }}{\\text{ mL }}=50\\text{ % }[\/latex]<\/p>\r\nSolutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. \u00a0In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.\r\n\r\nWe will use the following table to help us solve mixture problems:\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\"><\/td>\r\n<td class=\"border\"><\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nTo demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is [latex]120mL[\/latex] of a [latex]9\\%[\/latex] solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.\r\n\r\nFirst, find the total mass of solids for each solution by multiplying the volume by the concentration.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0[latex]120 mL[\/latex]<\/td>\r\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0[latex]75 mL[\/latex]<\/td>\r\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNext we add the new volumes and new masses.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0[latex]120 mL[\/latex]<\/td>\r\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0[latex]75 mL[\/latex]<\/td>\r\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]195 mL[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. \u00a0In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA chemist has [latex]70 mL[\/latex] of a [latex]50\\%[\/latex] methane solution. How much of an [latex]80\\%[\/latex] solution must she add so the final solution is [latex]60\\%[\/latex] methane?\r\n[reveal-answer q=\"274848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"274848\"]\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/apply-a-problem-solving-strategy-to-basic-word-problems\/\">Let's use the problem solving process outlined previously<\/a> to help us work through a solution to the problem.\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are looking for a new amount - in this case a volume - \u00a0based on the words \"how much\". \u00a0We know two starting \u00a0concentrations and the final concentration, as well as one volume.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>Solution 1 is the [latex]70 mL[\/latex] of [latex]50\\%[\/latex] methane and solution 2 is the unknown amount with [latex]80\\%[\/latex] methane. \u00a0We can call our unknown amount [latex]x[\/latex].\r\n\r\n<strong>Write and Solve: \u00a0<\/strong>Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0[latex]70[\/latex]<\/td>\r\n<td class=\"border\">[latex]0.5[\/latex]<\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0[latex]x[\/latex]<\/td>\r\n<td class=\"border\">\u00a0[latex]0.8[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<td class=\"border\" style=\"text-align: left\">[latex]0.6[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nMultiply amount by concentration\u00a0to get total,\u00a0be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex]Add the entries in the amount column to get final amount. The concentration for this amount is [latex]0.6[\/latex] because we want the final solution to be [latex]60\\%[\/latex] methane.\r\n<table class=\" undefined alignleft\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a070<\/td>\r\n<td class=\"border\">\u00a00.5<\/td>\r\n<td class=\"border\">\u00a035<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0x<\/td>\r\n<td class=\"border\">\u00a00.8<\/td>\r\n<td>\u00a00.8<em>x<\/em><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left\">\u00a070+x<\/td>\r\n<td class=\"border\" style=\"text-align: left\">0.6<\/td>\r\n<td class=\"border\">\u00a0[latex]42+0.6x[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nAdd the total mass for solution 1 and solution 2 to get the total mass for the [latex]60\\%[\/latex] solution. This is our equation for finding the unknown volume.\r\n\r\n[latex]35+0.8x=42+0.6x[\/latex]\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.8x=42+0.6x\\\\\\underline{-0.6x}\\,\\,\\,\\,\\,\\,\\,\\underline{-0.6x}\\\\35+0.2x=42\\\\\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Subtract [latex]35[\/latex] from both sides<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\r\n<p style=\"text-align: left\">Divide both sides by [latex]0.2[\/latex]<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]\r\n[latex]x=35[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]35mL[\/latex] of the\u00a0[latex]80\\%[\/latex] solution must be added to the original [latex]70mL[\/latex] of the\u00a0[latex]50\\%[\/latex] solution to gain a solution with a concentration of [latex]60\\%[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe above problem illustrates how we can use\u00a0the mixture table\u00a0to define\u00a0an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nA farmer has two types of milk, one that is [latex]24\\%[\/latex] butterfat and another which is [latex]18\\%[\/latex] butterfat. How much of each should he use to end up with [latex]42[\/latex] gallons of [latex]20\\%[\/latex] butterfat?\r\n[reveal-answer q=\"966963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"966963\"]\r\n\r\n<strong>Read and Understand:\u00a0<\/strong>We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is [latex]42[\/latex] gallons. There are two unknowns in this problem.\r\n\r\n<strong>Define and Translate:\u00a0<\/strong>We will call the unknown volume of the \u00a0[latex]24\\%[\/latex] solution [latex]x[\/latex], and the unknown volume of the [latex]18\\%[\/latex] solution [latex]y[\/latex].\r\n\r\n<strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0[latex]x[\/latex]<\/td>\r\n<td class=\"border\">[latex]0.24[\/latex]<\/td>\r\n<td class=\"border\"><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0[latex]y[\/latex]<\/td>\r\n<td class=\"border\">[latex]0.18[\/latex]<\/td>\r\n<td><\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left\">[latex]42[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: left\">[latex]0.2[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: center\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nFind the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from\r\n<table class=\" undefined\">\r\n<thead>\r\n<tr class=\"border\">\r\n<th><\/th>\r\n<th class=\"border\">Amount<\/th>\r\n<th class=\"border\">Concentration (%)<\/th>\r\n<th>Total Mass<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 1<\/td>\r\n<td class=\"border\">\u00a0[latex]x[\/latex]<\/td>\r\n<td class=\"border\">[latex]0.24[\/latex]<\/td>\r\n<td class=\"border\">[latex]0.24x[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\">Solution 2<\/td>\r\n<td class=\"border\">\u00a0[latex]y[\/latex]<\/td>\r\n<td class=\"border\">[latex]0.18[\/latex]<\/td>\r\n<td>[latex]0.18y[\/latex]<\/td>\r\n<\/tr>\r\n<tr class=\"border\">\r\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\r\n<td class=\"border\" style=\"text-align: left\">[latex]x+y=42[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: left\">[latex]0.2[\/latex]<\/td>\r\n<td class=\"border\" style=\"text-align: left\">[latex]8.4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWhen you sum the amount column you get\u00a0one equation: [latex]x+ y = 42[\/latex]\r\nWhen you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[\/latex]\r\n\r\nUse elimination to find a value for [latex]x[\/latex] and [latex]y[\/latex].\r\n\r\nMultiply the first equation by [latex]-0.18[\/latex]\r\n\r\n[latex]-0.18(x+y) = (42)(-0.18)[\/latex]\r\n\r\n[latex]-0.18x - 0.18y = -7.56[\/latex]\r\n\r\nNow our system of equations looks like this:\r\n\r\n[latex]-0.18x - 0.18y = -7.56[\/latex]\r\n\r\n[latex]0.24x + 0.18y = 8.4[\/latex]\r\n\r\nAdding the two equations together to eliminate the y terms gives this equation:\r\n\r\n[latex]0.06x = 8.4[\/latex]\r\n\r\nDivide by [latex]0.06[\/latex] on each side:\r\n\r\n[latex]x = 14[\/latex]\r\n\r\nNow substitute the value for [latex]x[\/latex] into one of the equations in order to solve for [latex]y[\/latex].\r\n\r\n[latex](14) + y = 42[\/latex]\r\n\r\n[latex]y = 28[\/latex]\r\n<h4>Answer<\/h4>\r\nThis can be interpreted as [latex]14[\/latex] gallons of [latex]24\\%[\/latex] butterfat milk added to [latex]28[\/latex] gallons of [latex]18\\%[\/latex] butterfat milk will give [latex]42[\/latex] gallons of [latex]20\\%[\/latex] butterfat milk.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.\r\n\r\nhttps:\/\/youtu.be\/4s5MCqphpKo\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]3501[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve mixture problems<\/li>\n<\/ul>\n<\/div>\n<h2 id=\"title1\">Write a system of linear equations representing a mixture problem, solve the system and interpret the results<\/h2>\n<p>One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. \u00a0A solution is a mixture of two or more different substances like water and salt or vinegar and oil. \u00a0Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand. \u00a0There are many other disciplines that use solutions as well.<\/p>\n<p>The concentration or strength of a liquid solution is often described\u00a0\u00a0as a percentage. \u00a0This number comes from the ratio of how much mass is in a specific volume of liquid. \u00a0For example if you have [latex]50[\/latex] grams of salt in a [latex]100mL[\/latex] of water you have a [latex]50\\%[\/latex] salt solution based on the following ratio:<\/p>\n<p style=\"text-align: center\">[latex]\\frac{50\\text{ grams }}{100\\text{ mL }}=0.50\\frac{\\text{ grams }}{\\text{ mL }}=50\\text{ % }[\/latex]<\/p>\n<p>Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths. \u00a0In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.<\/p>\n<p>We will use the following table to help us solve mixture problems:<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\"><\/td>\n<td class=\"border\"><\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is [latex]120mL[\/latex] of a [latex]9\\%[\/latex] solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.<\/p>\n<p>First, find the total mass of solids for each solution by multiplying the volume by the concentration.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0[latex]120 mL[\/latex]<\/td>\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0[latex]75 mL[\/latex]<\/td>\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Next we add the new volumes and new masses.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0[latex]120 mL[\/latex]<\/td>\n<td class=\"border\">0.09 [latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]\\left(120\\cancel{\\text{ mL}}\\right)\\left(0.09\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=10.8\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0[latex]75 mL[\/latex]<\/td>\n<td class=\"border\">0.23\u00a0[latex]\\frac{\\text{ grams }}{\\text{ mL }}[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]\\left(75\\cancel{\\text{ mL}}\\right)\\left(0.23\\frac{\\text{ grams }}{\\cancel{\\text{ mL }}}\\right)=17.25\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]195 mL[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]\\frac{28.05\\text{ grams }}{ 195 \\text{ mL }}=0.14=14\\text{ % }[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\">[latex]10.8\\text{ grams }+17.25\\text{ grams }=28.05\\text{ grams }[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution. \u00a0In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A chemist has [latex]70 mL[\/latex] of a [latex]50\\%[\/latex] methane solution. How much of an [latex]80\\%[\/latex] solution must she add so the final solution is [latex]60\\%[\/latex] methane?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q274848\">Show Solution<\/span><\/p>\n<div id=\"q274848\" class=\"hidden-answer\" style=\"display: none\">\n<p><a href=\"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/apply-a-problem-solving-strategy-to-basic-word-problems\/\">Let&#8217;s use the problem solving process outlined previously<\/a> to help us work through a solution to the problem.<\/p>\n<p><strong>Read and Understand:\u00a0<\/strong>We are looking for a new amount &#8211; in this case a volume &#8211; \u00a0based on the words &#8220;how much&#8221;. \u00a0We know two starting \u00a0concentrations and the final concentration, as well as one volume.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>Solution 1 is the [latex]70 mL[\/latex] of [latex]50\\%[\/latex] methane and solution 2 is the unknown amount with [latex]80\\%[\/latex] methane. \u00a0We can call our unknown amount [latex]x[\/latex].<\/p>\n<p><strong>Write and Solve: \u00a0<\/strong>Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0[latex]70[\/latex]<\/td>\n<td class=\"border\">[latex]0.5[\/latex]<\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0[latex]x[\/latex]<\/td>\n<td class=\"border\">\u00a0[latex]0.8[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<td class=\"border\" style=\"text-align: left\">[latex]0.6[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Multiply amount by concentration\u00a0to get total,\u00a0be sure to distribute on the last row: [latex]\\left(70 + x\\right)0.6[\/latex]Add the entries in the amount column to get final amount. The concentration for this amount is [latex]0.6[\/latex] because we want the final solution to be [latex]60\\%[\/latex] methane.<\/p>\n<table class=\"undefined alignleft\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a070<\/td>\n<td class=\"border\">\u00a00.5<\/td>\n<td class=\"border\">\u00a035<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0x<\/td>\n<td class=\"border\">\u00a00.8<\/td>\n<td>\u00a00.8<em>x<\/em><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left\">\u00a070+x<\/td>\n<td class=\"border\" style=\"text-align: left\">0.6<\/td>\n<td class=\"border\">\u00a0[latex]42+0.6x[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Add the total mass for solution 1 and solution 2 to get the total mass for the [latex]60\\%[\/latex] solution. This is our equation for finding the unknown volume.<\/p>\n<p>[latex]35+0.8x=42+0.6x[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.8x=42+0.6x\\\\\\underline{-0.6x}\\,\\,\\,\\,\\,\\,\\,\\underline{-0.6x}\\\\35+0.2x=42\\\\\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Subtract [latex]35[\/latex] from both sides<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}35+0.2x=42\\\\\\underline{-35}\\,\\,\\,\\,\\,\\,\\,\\underline{-35}\\\\0.2x=7\\end{array}[\/latex]<\/p>\n<p style=\"text-align: left\">Divide both sides by [latex]0.2[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{c}0.2x=7\\\\\\frac{0.2x}{0.2}=\\frac{7}{0.2}\\end{array}[\/latex]<br \/>\n[latex]x=35[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]35mL[\/latex] of the\u00a0[latex]80\\%[\/latex] solution must be added to the original [latex]70mL[\/latex] of the\u00a0[latex]50\\%[\/latex] solution to gain a solution with a concentration of [latex]60\\%[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The above problem illustrates how we can use\u00a0the mixture table\u00a0to define\u00a0an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>A farmer has two types of milk, one that is [latex]24\\%[\/latex] butterfat and another which is [latex]18\\%[\/latex] butterfat. How much of each should he use to end up with [latex]42[\/latex] gallons of [latex]20\\%[\/latex] butterfat?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q966963\">Show Solution<\/span><\/p>\n<div id=\"q966963\" class=\"hidden-answer\" style=\"display: none\">\n<p><strong>Read and Understand:\u00a0<\/strong>We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is [latex]42[\/latex] gallons. There are two unknowns in this problem.<\/p>\n<p><strong>Define and Translate:\u00a0<\/strong>We will call the unknown volume of the \u00a0[latex]24\\%[\/latex] solution [latex]x[\/latex], and the unknown volume of the [latex]18\\%[\/latex] solution [latex]y[\/latex].<\/p>\n<p><strong>Write and Solve:\u00a0<\/strong>Fill in the table with the information we know.<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0[latex]x[\/latex]<\/td>\n<td class=\"border\">[latex]0.24[\/latex]<\/td>\n<td class=\"border\"><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0[latex]y[\/latex]<\/td>\n<td class=\"border\">[latex]0.18[\/latex]<\/td>\n<td><\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left\">[latex]42[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: left\">[latex]0.2[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: center\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from<\/p>\n<table class=\"undefined\">\n<thead>\n<tr class=\"border\">\n<th><\/th>\n<th class=\"border\">Amount<\/th>\n<th class=\"border\">Concentration (%)<\/th>\n<th>Total Mass<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr class=\"border\">\n<td class=\"border\">Solution 1<\/td>\n<td class=\"border\">\u00a0[latex]x[\/latex]<\/td>\n<td class=\"border\">[latex]0.24[\/latex]<\/td>\n<td class=\"border\">[latex]0.24x[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\">Solution 2<\/td>\n<td class=\"border\">\u00a0[latex]y[\/latex]<\/td>\n<td class=\"border\">[latex]0.18[\/latex]<\/td>\n<td>[latex]0.18y[\/latex]<\/td>\n<\/tr>\n<tr class=\"border\">\n<td class=\"border\" style=\"text-align: left\">Final Solution<\/td>\n<td class=\"border\" style=\"text-align: left\">[latex]x+y=42[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: left\">[latex]0.2[\/latex]<\/td>\n<td class=\"border\" style=\"text-align: left\">[latex]8.4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>When you sum the amount column you get\u00a0one equation: [latex]x+ y = 42[\/latex]<br \/>\nWhen you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[\/latex]<\/p>\n<p>Use elimination to find a value for [latex]x[\/latex] and [latex]y[\/latex].<\/p>\n<p>Multiply the first equation by [latex]-0.18[\/latex]<\/p>\n<p>[latex]-0.18(x+y) = (42)(-0.18)[\/latex]<\/p>\n<p>[latex]-0.18x - 0.18y = -7.56[\/latex]<\/p>\n<p>Now our system of equations looks like this:<\/p>\n<p>[latex]-0.18x - 0.18y = -7.56[\/latex]<\/p>\n<p>[latex]0.24x + 0.18y = 8.4[\/latex]<\/p>\n<p>Adding the two equations together to eliminate the y terms gives this equation:<\/p>\n<p>[latex]0.06x = 8.4[\/latex]<\/p>\n<p>Divide by [latex]0.06[\/latex] on each side:<\/p>\n<p>[latex]x = 14[\/latex]<\/p>\n<p>Now substitute the value for [latex]x[\/latex] into one of the equations in order to solve for [latex]y[\/latex].<\/p>\n<p>[latex](14) + y = 42[\/latex]<\/p>\n<p>[latex]y = 28[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>This can be interpreted as [latex]14[\/latex] gallons of [latex]24\\%[\/latex] butterfat milk added to [latex]28[\/latex] gallons of [latex]18\\%[\/latex] butterfat milk will give [latex]42[\/latex] gallons of [latex]20\\%[\/latex] butterfat milk.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex:  System of Equations Application - Mixture Problem\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4s5MCqphpKo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm3501\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=3501&theme=oea&iframe_resize_id=ohm3501&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-16572\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex: System of Equations Application - Mixture Problem. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4s5MCqphpKo\">https:\/\/youtu.be\/4s5MCqphpKo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Beginning and Intermediate Algebra Textbook. <strong>Authored by<\/strong>: Tyler Wallace. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.wallace.ccfaculty.org\/book\/book.html\">http:\/\/www.wallace.ccfaculty.org\/book\/book.html<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstaxcollege.org\/textbooks\/college-algebra\">https:\/\/openstaxcollege.org\/textbooks\/college-algebra<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":169554,"menu_order":12,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Beginning and Intermediate Algebra Textbook\",\"author\":\"Tyler Wallace\",\"organization\":\"\",\"url\":\"www.wallace.ccfaculty.org\/book\/book.html\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc-attribution\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstaxcollege.org\/textbooks\/college-algebra\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: System of Equations Application - Mixture Problem\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/4s5MCqphpKo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 14: Systems of Equations and Inequalities, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\" http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"0b22215a478e41e292cc1317adbff940, 5cbe51bf24f143b09646f7e3d8a1ec55","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16572","chapter","type-chapter","status-publish","hentry"],"part":16192,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16572","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16572\/revisions"}],"predecessor-version":[{"id":20426,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16572\/revisions\/20426"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/16192"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16572\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16572"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16572"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16572"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/suny-rockland-developmentalemporium\/wp-json\/wp\/v2\/license?post=16572"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}