{"id":6744,"date":"2020-10-08T17:03:51","date_gmt":"2020-10-08T17:03:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6744"},"modified":"2026-02-01T05:23:23","modified_gmt":"2026-02-01T05:23:23","slug":"4-3-solving-a-2x2-system-of-linear-equations-by-elimination","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/4-3-solving-a-2x2-system-of-linear-equations-by-elimination\/","title":{"raw":"4.3: Solving a 2x2 System of Linear Equations by Elimination","rendered":"4.3: Solving a 2×2 System of Linear Equations by Elimination"},"content":{"raw":"
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Section 4.3 Learning Objectives<\/h3>\r\n
\r\n
<\/div>\r\n
\r\n\r\n4.3:\u00a0 Solving a 2x2 System of Linear Equations by Addition\/Elimination<\/strong>\r\n\r\n<\/div>\r\n
\r\n
    \r\n \t
  • Solve systems of linear equations using elimination<\/li>\r\n \t
  • Recognize when systems of linear equations have no solution or an infinite number of solutions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n
    \r\n
    <\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n

    Solve a system of equations using the elimination\u00a0method<\/h2>\r\nThe elimination method<\/b> for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation to eliminate one of the variable terms. \u00a0In this method, you may or may not need to multiply the terms in one equation by a number first. \u00a0We will first look at examples where no multiplication is necessary to use the elimination method.\u00a0 Then you will see examples using multiplication after you are familiar with the idea of the elimination method.\r\n\r\nIt is easier to show rather than tell with this method, so let's dive right into some examples.\r\n\r\nIf you add the two equations,\r\n\r\n[latex]x\u2013y=\u22126[\/latex] and [latex]x+y=8[\/latex] together, watch what happens.\r\n

    [latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,x-y=\\,-6\\\\\\underline{+\\,x+y=\\,\\,\\,8}\\\\\\,2x+0\\,=\\,\\,\\,\\,2\\end{array}[\/latex]<\/p>\r\nYou have eliminated the [latex]y [\/latex]-term, and this equation can be solved using the methods for solving equations with one variable.\r\n\r\nLet\u2019s see how this system is solved using the elimination method.\r\n

    \r\n

    Example 1<\/h3>\r\nUse elimination to solve the system.\r\n

    [latex]\\begin{array}{r}x\u2013y=\u22126\\\\x+y=\\,\\,\\,\\,8\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"403819\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"403819\"]Add the equations.\r\n

    [latex] \\displaystyle \\begin{array}{r}x-y=\\,\\,-6\\\\+\\underline{\\,x\\,+y=\\,\\,\\,\\,\\,8}\\\\\\,\\,\\,2x\\,\\,\\,\\,\\,\\,\\,\\,=\\,\\,\\,\\,2\\,\\,\\end{array}[\/latex]<\/p>\r\nSolve for [latex]x[\/latex].\r\n

    [latex]\\begin{array}{r}\\displaystyle\\frac{2x}{2}=\\frac{2}{2}\\\\x=1\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]x=1[\/latex] into one of the original equations and solve for y<\/i>.\r\n

    [latex]\\begin{array}{r}x+y=8\\hspace{.05in}\\\\1+y=8\\hspace{.05in}\\\\ \\underline{-1\\hspace{.35in}-1}\\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=7\\end{array}[\/latex]<\/p>\r\nBe sure to check your answer in both equations!\r\n

    [latex]\\begin{array}{r}x\u2013y=\u22126\\\\1\u20137=\u22126\\\\\u22126=\u22126\\\\\\text{TRUE}\\\\\\\\x+y=8\\\\1+7=8\\\\8=8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n

    Answer<\/h4>\r\nThe solution is (1, 7).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nUnfortunately not all systems work out this easily. How about a system like [latex]2x+y=12[\/latex] and [latex]\u22123x+y=2[\/latex]. If you add these two equations together, no variables are eliminated.\r\n

    [latex] \\displaystyle \\begin{array}{l}\\,\\,\\,\\,2x+y=12\\\\\\underline{-3x+y=\\,\\,\\,2}\\\\-x+2y=14\\end{array}[\/latex]<\/p>\r\nBut you want to eliminate a variable. So let\u2019s add the opposite of one of the equations to the other equation. This means multiply every term in one of the equations by [latex]-1 [\/latex], so that the sign of every term is opposite.\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,2x+\\,\\,y\\,=12\\rightarrow2x+y=12\\rightarrow2x+y=12\\\\\u22123x+\\,\\,y\\,=2\\rightarrow\u2212\\left(\u22123x+y\\right)=\u2212(2)\\rightarrow3x\u2013y=\u22122\\\\\\,\\,\\,\\,5x+0y=10\\end{array}[\/latex]<\/p>\r\nYou have eliminated the [latex]y[\/latex]-variable, and the problem can now be solved.\u00a0 The complete solution is shown in the example after the video.\r\n\r\nThe following video describe a similar problem where you can eliminate one variable by adding the two equations together.\r\n\r\nhttps:\/\/youtu.be\/M4IEmwcqR3c\r\n

    \"Caution\"Caution! \u00a0When you add the opposite of one entire equation to another, make sure to change the sign of EVERY term on both sides of the equation. This is a very common\u00a0mistake to make.<\/div>\r\n
    \r\n

    Example 2<\/h3>\r\nUse elimination to solve the system.\r\n

    [latex]\\begin{array}{r}2x+y=12\\\\\u22123x+y=2\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"702178\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"702178\"]You can eliminate the [latex]y [\/latex]-variable if you add the opposite of one of the equations to the other equation.\r\n

    [latex]\\begin{array}{r}2x+y=12\\\\\u22123x+y=2\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nRewrite the second equation as its opposite. In other words, multiply by [latex]-1[\/latex].\r\n\r\nAdd.\u00a0Solve for [latex]x [\/latex].\r\n

    [latex]\\begin{array}{r}2x+y=12\\hspace{.05in}\\\\\\underline{3x\\,\\,\u2013\\,y=\u22122}\\\\5x\\hspace{.3in}=\\,10\\hspace{.05in}\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.26in}\\displaystyle\\frac{5x}{5}=\\frac{10}{5}[\/latex]<\/p>\r\n

    [latex]\\hspace{.26in}x=2[\/latex]<\/p>\r\nSubstitute [latex]x=2[\/latex] into one of the original equations and solve for [latex]y [\/latex].\r\n

    [latex]\\begin{array}{r}2\\left(2\\right)+y=12\\\\4+y=12\\\\\\underline{-4\\hspace{.4in}-4}\\\\y=8\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nBe sure to check your answer in both equations!\r\n

    [latex]\\begin{array}{r}2x+y=12\\\\2\\left(2\\right)+8=12\\\\4+8=12\\\\12=12\\\\\\text{TRUE}\\\\\\\\\u22123x+y=2\\\\\u22123\\left(2\\right)+8=2\\\\\u22126+8=2\\\\2=2\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n

    Answer<\/h4>\r\nThe solution is (2, 8).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following are two more examples showing how to solve linear systems of equations using elimination.\r\n
    \r\n

    Example 3<\/h3>\r\nUse elimination to solve the system.\r\n

    [latex]\\begin{array}{r}3y=2x-1 \\\\ 2x=5(5-y)\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"438400\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"438400\"]First, note that we will find the elimination process easier if the equations are in the standard form, [latex]Ax+By=C[\/latex]. So, we will start by subtracting [latex]2x[\/latex] from both sides in the first equation.\r\n

    [latex]3y=2x-1[\/latex]<\/p>\r\n

    [latex]-2x+3y=-1[\/latex]<\/p>\r\nThen we will distribute and add [latex]5y[\/latex] to both sides in the second equation.\r\n

    [latex]2x=5(5-y)[\/latex]<\/p>\r\n

    [latex]2x=25-5y[\/latex]<\/p>\r\n

    [latex]2x+5y=25[\/latex]<\/p>\r\nThis results in the system shown below.\r\n

    [latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\2x+5y=\\,25\\hspace{.02in}\\end{array}[\/latex]<\/p>\r\nNotice the coefficients of each variable in each equation. If you add these two equations, the [latex]x [\/latex]-term will be eliminated since [latex]\u22122x+2x=0[\/latex]. Thus, we can add the equations and solve for [latex]y [\/latex].\r\n

    [latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\\\underline{2x+5y=25\\,}\\\\8y=24\\,\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.49in}\\dfrac{8y}{8}=\\dfrac{24}{8}[\/latex]<\/p>\r\n

    [latex]\\hspace{.49in}y=3[\/latex]<\/p>\r\nSubstitute [latex]y=3[\/latex] <\/i>into one of the original equations.\r\n

    [latex]\\begin{array}{r}2x+5y=25\\\\2x+5\\left(3\\right)=25\\\\2x+15=25\\\\\\underline{\\hspace{.2in}-\\hspace{.03in}15\\hspace{.03in}-\\hspace{-.01in}15}\\\\2x\\hspace{.38in}=10\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.58in}\\dfrac{2x}{2}=\\dfrac{10}{2}[\/latex]<\/p>\r\n

    [latex]\\hspace{.58in}x=5[\/latex]<\/p>\r\nCheck solutions.\r\n

    [latex]\\begin{array}{r}\u22122x+3y=\u22121\\\\\u22122\\left(5\\right)+3\\left(3\\right)=\u22121\\\\\u221210+9=\u22121\\\\\u22121=\u22121\\\\\\text{TRUE}\\\\\\\\2x+5y=25\\\\2\\left(5\\right)+5\\left(3\\right)=25\\\\10+15=25\\\\25=25\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n

    Answer<\/h4>\r\nThe solution is (5, 3).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \r\n

    Example 4<\/h3>\r\nUse elimination to solve for\u00a0 [latex]x[\/latex]\u00a0<\/i>and\u00a0 [latex]y[\/latex].<\/i>\r\n

    [latex]\\begin{array}{r}4x+2y=14\\\\5x+2y=16\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"776093\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"776093\"]Notice the coefficients of each variable in each equation. You will need to add the opposite of one of the equations to eliminate the variable y<\/i>, as [latex]2y+2y=4y[\/latex], but [latex]2y+\\left(\u22122y\\right)=0[\/latex].\r\n\r\nChange one of the equations to its opposite (i.e. multiply by [latex]-1[\/latex]), add, and solve for [latex]x[\/latex]. Let's multiply the second equation by [latex]-1[\/latex].\r\n

    [latex]\\begin{array}{r}4x+2y=14\\,\\,\\,\\,\\,\\\\\\underline{\u22125x\\hspace{.06in}\u2013\\hspace{.11in}2y=\u221216}\\\\\u2212x\\hspace{.4in}=\u22122\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.37in}\\dfrac{-x}{-1}=\\dfrac{-2}{-1} [\/latex]<\/p>\r\n

    [latex]\\hspace{.37in}x=2[\/latex]<\/p>\r\nSubstitute [latex]x=2[\/latex] into one of the original equations and solve for [latex]y [\/latex].\r\n

    [latex]\\begin{array}{r}4x+2y=14\\\\4\\left(2\\right)+2y=14\\\\8+2y=14\\\\\\underline{\\hspace{.02in}-8\\hspace{.49in}-8}\\\\2y=6\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.47in}\\dfrac{2y}{2}=\\dfrac{6}{2} [\/latex]<\/p>\r\n

    [latex]\\hspace{.55in}y=3[\/latex]<\/p>\r\nWe will leave it to you to check this solution.\r\n

    Answer<\/h4>\r\nThe solution is (2, 3).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGo ahead and check this last example\u2014substitute [latex](2, 3)[\/latex] into both equations. You get two true statements: [latex]14=14[\/latex] and [latex]16=16[\/latex].\r\n\r\nNotice that you could have used the opposite of the first equation rather than the second equation and the result would be the same.\r\n

    Recognize systems that have no solution or an infinite number of solutions<\/h2>\r\nJust as with the substitution method, the elimination method will sometimes eliminate both v<\/i>ariables, and you end up with either a true statement or a false statement. Recall that a false statement means that there is no solution.\r\n\r\nLet\u2019s look at an example.\r\n
    \r\n

    Example 5<\/h3>\r\nSolve for\u00a0 [latex]x[\/latex]\u00a0<\/i>and\u00a0 [latex]y [\/latex].<\/i>\r\n

    [latex]\\begin{array}{r}-\\hspace{.02in}x\\hspace{.03in}\u2013\\hspace{.03in}y=-4\\\\x+y=2\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"101540\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"101540\"]Add the equations to eliminate the\u00a0x<\/em>-term.\r\n

    [latex]\\begin{array}{r}-\\hspace{.02in}x\\hspace{.03in}\u2013\\hspace{.03in}y=-4\\\\\\underline{x+y=2\\,\\,\\,\\,}\\\\0=\u22122\\end{array}[\/latex]<\/p>\r\nThis false statement leads us to conclude that the system has no solution.\r\n

    Answer<\/h4>\r\nNo Solution\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGraphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that there is no solution.\r\n\r\n\"Two\r\n\r\nIf both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.\r\n
    \r\n

    Example 6<\/h3>\r\nSolve for\u00a0 [latex]x[\/latex]\u00a0<\/i>and [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}x+y=2\\,\\,\\,\\,\\,\\\\-x\u2212y=-2\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"328100\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"328100\"]Add the equations to eliminate the\u00a0x<\/i>-term.\r\n

    [latex]\\begin{array}{r}x+y=2\\,\\,\\,\\,\\,\\\\\\underline{-x\u2212y=-2}\\\\0=0\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nThis true statement leads us to conclude that there are an infinite number of solutions, given by all points on the line. Recall that we learned the proper way of expressing such an answer earlier in this chapter.\r\n

    Answer<\/h4>\r\nThere are an infinite number of solutions, given by\r\n\r\n[latex]\\{(x,y)\\hspace{.01in}|\\hspace{.01in}x+y=2\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nGraphing these two equations will help to illustrate what is happening.\r\n\r\n\"Coordinate<\/b>\r\n

    <\/h2>\r\nIn the following video, a system of equations which has no solutions is solved using the method of elimination.\r\n\r\nhttps:\/\/youtu.be\/z5_ACYtzW98\r\n

    Solve a system of equations when multiplication is necessary to eliminate a variable<\/h2>\r\nMany times adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\nIf you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let\u2019s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will allow you to eliminate\u00a0the same variable in the other equation.\r\n\r\nWe do this with multiplication.\u00a0\u00a0Notice that the first equation contains the term [latex]4y [\/latex], and the second equation contains the term y<\/i>. If you multiply the second equation by [latex]\u22124 [\/latex], when you add both equations the\u00a0 [latex]y [\/latex]-<\/i>variable terms will add up to [latex]0 [\/latex].\r\n\r\nThe following example takes you through all the steps to find a solution to this system.\r\n
    \r\n

    Example 7<\/h3>\r\nSolve for [latex]x[\/latex]and [latex]y[\/latex].\r\n\r\n[latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"815377\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815377\"]Look for terms that can be eliminated. In this case, neither the\u00a0x<\/i> or y\u00a0<\/i>terms will cancel.\r\n

    [latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\r\nMultiply the second equation by [latex]\u22124[\/latex] so they do have the same coefficient.\r\n

    [latex]\u22124\\left(5x+y\\right)=\u22124\\left(30\\right)[\/latex]<\/p>\r\nThis gives\r\n

    [latex]-20x-4y=-120[\/latex]<\/p>\r\nRewrite the system and add the equations.\r\n

    [latex]\\begin{array}{r}3x+4y=52\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{\u221220x\\hspace{.04in}\u2013\\hspace{.03in}4y=\u2212120}\\\\-17x\\hspace{.36in}=-68\\hspace{.12in}\\end{array}[\/latex]<\/p>\r\nSolve for x<\/i>.\r\n

    [latex]\\dfrac{\u221217x}{-17}=\\dfrac{-68}{-17}[\/latex]<\/p>\r\n

    \u00a0[latex]\\hspace{.05in}x=4[\/latex]<\/p>\r\nSubstitute [latex]x=4[\/latex] into one of the original equations to find y<\/i>.\r\n

    [latex]\\begin{array}{r}3x+4y=52\\,\\\\3\\left(4\\right)+4y=52\\,\\\\12+4y=52\\,\\\\\\underline{-12\\hspace{.43in}-12}\\\\4y=40\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.54in}\\dfrac{4y}{4}=\\dfrac{40}{4}[\/latex]<\/p>\r\n

    [latex]\\hspace{.6in}y=10[\/latex]<\/p>\r\nCheck your answer.\r\n

    [latex]\\begin{array}{r}3x+4y=52\\\\3\\left(4\\right)+4\\left(10\\right)=52\\\\12+40=52\\\\52=52\\\\\\text{TRUE}\\\\\\\\5x+y=30\\\\5\\left(4\\right)+10=30\\\\20+10=30\\\\30=30\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\nThe answers check.\r\n

    Answer<\/h4>\r\nThe solution is (4, 10).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
    \"CautionCaution! \u00a0When you\u00a0use multiplication to eliminate a variable, you must multiply EACH term in the equation by the number you choose. \u00a0Forgetting to multiply every term is a common mistake.<\/div>\r\nThere are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied both<\/i> equations by different numbers.\r\n\r\nLet\u2019s eliminate the variable\u00a0 [latex]x [\/latex]\u00a0<\/i>this time. We can achieve this by multiplying the first equation by 5 and the second equation by [latex]\u22123[\/latex].\r\n
    \r\n

    Example 8<\/h3>\r\nSolve for [latex]x[\/latex] and [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"40585\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"40585\"]Look for terms that can be eliminated. The equations do not have any\u00a0[latex]x[\/latex] or [latex]y[\/latex]terms with opposite coefficients.\r\n

    [latex]\\begin{array}{r}3x+4y=52\\\\5x+y=30\\end{array}[\/latex]<\/p>\r\nIn order to use the elimination method, you have to create variables that have the same coefficient in absolute value, but opposite signs\u2014then you can eliminate them by adding the equation. Multiply the top equation by 5.\r\n

    [latex]\\begin{array}{r}5\\left(3x+4y\\right)=5\\left(52\\right)\\\\15x+20y=260\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nNow multiply the bottom equation by \u22123.\r\n

    [latex]\\begin{array}{r}-3(5x+y)=\u22123(30)\\\\\u221215x\u20133y=\u221290\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nNext add the equations, and solve for [latex]y[\/latex].\r\n

    [latex]\\begin{array}{r}15x+20y=260\\hspace{.04in}\\\\\\underline{\u221215x\\hspace{.09in}\u2013\\hspace{.11in}3y=\\,\u201390}\\\\17y=170\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.62in}\\dfrac{17y}{17}=\\dfrac{170}{17}\\,[\/latex]<\/p>\r\n

    [latex]\\hspace{.67in}y=10[\/latex]<\/p>\r\nSubstitute [latex]y=10[\/latex] into one of the original equations to find x<\/i>.\r\n

    [latex]\\begin{array}{r}3x+4y=52\\,\\\\3x+4\\left(10\\right)=52\\,\\\\3x+40=52\\,\\\\\\underline{\\hspace{.2in}-\\hspace{.05in}40\\hspace{.04in}-\\hspace{-.02in}40}\\\\3x=12\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.6in}\\dfrac{3x}{3}=\\dfrac{12}{3}[\/latex]<\/p>\r\n

    [latex]\\hspace{.6in}x=4[\/latex]<\/p>\r\nYou arrive at the same solution as before.\r\n

    Answer<\/span><\/h4>\r\nThe solution is (4, 10).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThese equations were multiplied by 5 and [latex]\u22123[\/latex] respectively, because that gave you terms that would add up to 0. Be sure to multiply all of the terms of the equation.\r\n\r\nIn the following video, you will see an example of using the elimination method for solving a system of equations.\r\n\r\nhttps:\/\/youtu.be\/_liDhKops2w\r\n\r\nThe next example reminds us that sometimes we must deal with fractions.\r\n
    \r\n

    Example 9<\/h3>\r\nUse elimination to solve the system of equations.\r\n\r\n[latex]\\begin{array}{l}\\frac{1}{2}x-\\frac{3}{4}y=1\\\\ \\frac{3}{4}x-\\frac{3}{2}y=\\frac{13}{8} \\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"797725\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"797725\"]\r\n\r\nWhile we could try to jump into the elimination process immediately, you may find it easier to first clear out the fractions by multiplying each equation by its LCD.\r\n\r\nMultiply the first equation by 4.\r\n

    [latex]4\\cdot \\frac{1}{2}x-4\\cdot \\frac{3}{4}y=4\\cdot 1[\/latex]<\/p>\r\n

    [latex]2x-3y=4[\/latex]<\/p>\r\nAnd multiply the second equation by 8.\r\n

    [latex]8\\cdot \\frac{3}{4}x-8\\cdot \\frac{3}{2}y=8\\cdot \\frac{13}{8}[\/latex]<\/p>\r\n

    [latex]6x-12y=13[\/latex]<\/p>\r\nSo, we are trying to solve the equivalent system below.\r\n

    [latex]\\begin{array}{r}2x-3y=4\\,\\,\\,\\\\6x-12y=13\\end{array}[\/latex]<\/p>\r\nUsing elimination, we can multiply the top equation by [latex]-3[\/latex] and add.\r\n

    [latex]\\begin{array}{r}-6x+\\hspace{.1in}9y=-12\\\\\\underline{6x-12y=13\\,}\\,\\,\\,\\\\-3y=1\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.43in}\\dfrac{-3y}{-3}=\\dfrac{1}{-3}[\/latex]<\/p>\r\n

    [latex]\\hspace{.66in} y=-\\frac{1}{3}[\/latex]<\/p>\r\nSo, coincidentally, we also obtain a fractional answer. We now plug this value in to find\u00a0[latex]x[\/latex].\u00a0<\/em>For simplicity, we can plug into one of the two equations we obtained after eliminating the fractions.\r\n

    [latex]2x-3y=4\\hspace{.1in}[\/latex]<\/p>\r\n

    [latex]2x-3\\left(-\\frac{1}{3}\\right)=4\\hspace{.52in}[\/latex]<\/p>\r\n

    [latex]\\begin{array}{r}2x+1=4\\,\\,\\\\\\underline{\\hspace{.24in}-\\hspace{.03in}1\\hspace{.04in}-1}\\\\2x\\hspace{.3in}=3\\,\\,\\end{array}[\/latex]<\/p>\r\n

    [latex]\\hspace{.25in}\\dfrac{2x}{3}=\\dfrac{3}{2}[\/latex]<\/p>\r\n

    [latex]\\hspace{.39in}x=\\dfrac{3}{2}[\/latex]<\/p>\r\nWe conclude that the solution to the system is the ordered pair [latex]\\left(\\frac{3}{2},-\\frac{1}{3}\\right)[\/latex].\r\n

    Answer<\/span><\/h4>\r\nThe solution is\u00a0[latex]\\left(\\frac{3}{2},-\\frac{1}{3}\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n \r\n\r\n<\/div>\r\n \r\n\r\nIt is possible to use the elimination method with multiplication and get a result that indicates no solutions or infinitely many solutions, just as we saw in simpler systems earlier. In the following example, you will see a system that has infinitely many solutions.\r\n
    \r\n

    Example 10<\/h3>\r\nSolve the system of equations.\r\n\r\n[latex]\\begin{array}{r}x-3y=-2\\\\-2x+6y=4\\end{array}[\/latex]\r\n\r\n[reveal-answer q=\"815327\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"815327\"]Look for terms that can be eliminated. The equations do not have any [latex]x[\/latex] or [latex]y[\/latex] terms with opposite coefficients.\r\n

    [latex]\\begin{array}{r}x-3y=-2\\\\-2x+6y=4\\end{array}[\/latex]<\/p>\r\nMultiply the first equation by [latex]2[\/latex] so that the [latex]x[\/latex] terms will cancel out.\r\n

    [latex]2\\left(x-3y\\right)=2(-2)[\/latex]<\/p>\r\n

    [latex]2x-6y=-4[\/latex]<\/p>\r\nRewrite the system and add the equations.\r\n

    [latex]\\begin{array}{r}2x-6y=-4\\\\\\underline{-2x+6y=4}\\hspace{.14in}\\\\0=0\\hspace{.14in}\\end{array}[\/latex]<\/p>\r\nDoes this kind of solution\u00a0look familiar? \u00a0This implied that there were an infinite number of solution. If we solve both of these equations for [latex]y[\/latex], you will see that they are the same equation.\r\n\r\nSolve the first equation for [latex]y[\/latex]:\r\n

    [latex]\\begin{array}{r}x-3y=-2\\\\-3y=-x-2\\\\y=\\frac{1}{3}x+\\frac{2}{3}\\end{array}[\/latex]<\/p>\r\nSolve the second equation for [latex]y[\/latex]:\r\n

    [latex]\\begin{array}{r} -2x+6y=4\\\\6y=2x+4\\\\y=\\frac{1}{3}x+\\frac{2}{3}\\end{array}[\/latex]<\/p>\r\nBoth equations are the same when written in slope intercept form, and therefore the solution set consists of all ordered pairs that lie on the line.\r\n

    Answer<\/h4>\r\nThere are an infinite number of solutions, given by\r\n\r\n[latex]\\{(x,y)\\hspace{.01in}|\\hspace{.01in}x-3y=-2\\}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, the elimination method is used to solve a system of equations. Notice that one of the equations needs to be multiplied by a\u00a0negative one first. \u00a0Additionally, this system has an infinite number of solutions.\r\n\r\nhttps:\/\/youtu.be\/NRxh9Q16Ulk\r\n

    Summary<\/h2>\r\nCombining equations is a powerful tool for solving a system of equations. Adding or subtracting two equations in order to eliminate a common variable is called the elimination (or addition) method. Once one variable is eliminated, it becomes much easier to solve for the other one.\r\n\r\nMultiplication can be used to set up terms that can be eliminated in the equations before they are combined to aid in finding a solution to a system. When using the multiplication method, it is important to multiply all the terms on both sides of the equation\u2014not just the one term you are trying to eliminate.\u00a0Solving using the elimination method will yield one of three results: a single value for each variable within the system (indicating one solution), an untrue statement (indicating no solutions), or a true statement (indicating an infinite number of solutions).<\/span>","rendered":"
    \n

    Section 4.3 Learning Objectives<\/h3>\n
    \n
    <\/div>\n
    \n

    4.3:\u00a0 Solving a 2×2 System of Linear Equations by Addition\/Elimination<\/strong><\/p>\n<\/div>\n

    \n
      \n
    • Solve systems of linear equations using elimination<\/li>\n
    • Recognize when systems of linear equations have no solution or an infinite number of solutions<\/li>\n<\/ul>\n<\/div>\n<\/div>\n
      \n
      <\/div>\n<\/div>\n<\/div>\n

       <\/p>\n

      Solve a system of equations using the elimination\u00a0method<\/h2>\n

      The elimination method<\/b> for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation to eliminate one of the variable terms. \u00a0In this method, you may or may not need to multiply the terms in one equation by a number first. \u00a0We will first look at examples where no multiplication is necessary to use the elimination method.\u00a0 Then you will see examples using multiplication after you are familiar with the idea of the elimination method.<\/p>\n

      It is easier to show rather than tell with this method, so let’s dive right into some examples.<\/p>\n

      If you add the two equations,<\/p>\n

      [latex]x\u2013y=\u22126[\/latex] and [latex]x+y=8[\/latex] together, watch what happens.<\/p>\n

      [latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,\\,x-y=\\,-6\\\\\\underline{+\\,x+y=\\,\\,\\,8}\\\\\\,2x+0\\,=\\,\\,\\,\\,2\\end{array}[\/latex]<\/p>\n

      You have eliminated the [latex]y[\/latex]-term, and this equation can be solved using the methods for solving equations with one variable.<\/p>\n

      Let\u2019s see how this system is solved using the elimination method.<\/p>\n

      \n

      Example 1<\/h3>\n

      Use elimination to solve the system.<\/p>\n

      [latex]\\begin{array}{r}x\u2013y=\u22126\\\\x+y=\\,\\,\\,\\,8\\end{array}[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      Add the equations.<\/p>\n

      [latex]\\displaystyle \\begin{array}{r}x-y=\\,\\,-6\\\\+\\underline{\\,x\\,+y=\\,\\,\\,\\,\\,8}\\\\\\,\\,\\,2x\\,\\,\\,\\,\\,\\,\\,\\,=\\,\\,\\,\\,2\\,\\,\\end{array}[\/latex]<\/p>\n

      Solve for [latex]x[\/latex].<\/p>\n

      [latex]\\begin{array}{r}\\displaystyle\\frac{2x}{2}=\\frac{2}{2}\\\\x=1\\,\\,\\,\\end{array}[\/latex]<\/p>\n

      Substitute [latex]x=1[\/latex] into one of the original equations and solve for y<\/i>.<\/p>\n

      [latex]\\begin{array}{r}x+y=8\\hspace{.05in}\\\\1+y=8\\hspace{.05in}\\\\ \\underline{-1\\hspace{.35in}-1}\\\\ \\,\\,\\,\\,\\,\\,\\,\\,\\,\\,y=7\\end{array}[\/latex]<\/p>\n

      Be sure to check your answer in both equations!<\/p>\n

      [latex]\\begin{array}{r}x\u2013y=\u22126\\\\1\u20137=\u22126\\\\\u22126=\u22126\\\\\\text{TRUE}\\\\\\\\x+y=8\\\\1+7=8\\\\8=8\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\n

      The answers check.<\/p>\n

      Answer<\/h4>\n

      The solution is (1, 7).<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      Unfortunately not all systems work out this easily. How about a system like [latex]2x+y=12[\/latex] and [latex]\u22123x+y=2[\/latex]. If you add these two equations together, no variables are eliminated.<\/p>\n

      [latex]\\displaystyle \\begin{array}{l}\\,\\,\\,\\,2x+y=12\\\\\\underline{-3x+y=\\,\\,\\,2}\\\\-x+2y=14\\end{array}[\/latex]<\/p>\n

      But you want to eliminate a variable. So let\u2019s add the opposite of one of the equations to the other equation. This means multiply every term in one of the equations by [latex]-1[\/latex], so that the sign of every term is opposite.<\/p>\n

      [latex]\\begin{array}{l}\\,\\,\\,\\,2x+\\,\\,y\\,=12\\rightarrow2x+y=12\\rightarrow2x+y=12\\\\\u22123x+\\,\\,y\\,=2\\rightarrow\u2212\\left(\u22123x+y\\right)=\u2212(2)\\rightarrow3x\u2013y=\u22122\\\\\\,\\,\\,\\,5x+0y=10\\end{array}[\/latex]<\/p>\n

      You have eliminated the [latex]y[\/latex]-variable, and the problem can now be solved.\u00a0 The complete solution is shown in the example after the video.<\/p>\n

      The following video describe a similar problem where you can eliminate one variable by adding the two equations together.<\/p>\n