{"id":956,"date":"2021-10-08T18:56:38","date_gmt":"2021-10-08T18:56:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&p=956"},"modified":"2021-12-16T00:41:51","modified_gmt":"2021-12-16T00:41:51","slug":"7-2-the-substitution-method","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/7-2-the-substitution-method\/","title":{"raw":"7.2: The Substitution Method","rendered":"7.2: The Substitution Method"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Use the substitution method to solve systems of linear equations<\/li>\r\n \t
  • Express the solution of an inconsistent system of equations containing two variables<\/li>\r\n \t
  • Express the solution of a dependent system of equations containing two variables<\/li>\r\n<\/ul>\r\n<\/div>\r\n
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    Key words<\/h3>\r\n
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    • Contradiction<\/strong>: a statement that is always false<\/li>\r\n \t
    • Identity<\/strong>: a statement that is always true<\/li>\r\n<\/ul>\r\n<\/div>\r\n

      The Substitution Method<\/h2>\r\nIn this section we will learn the\u00a0substitution\u00a0method<\/em>\u00a0<\/strong>for finding a solution to a system of linear equations in two variables. We have used substitution to verify the solution of an equation, when we substituted values in for the [latex]x[\/latex] and [latex]y[\/latex] variables. The idea is similar when applied to solving systems.\r\n\r\nFirst we make the assumption that a system of equations has a solution. This means that the lines cross at a single point, which is turn means that at that point, the\u00a0[latex]x[\/latex]-values and the [latex]y[\/latex]-values in each equation are the same. This allows us to substitute either an\u00a0[latex]x[\/latex]-value or a [latex]y[\/latex]-value from one equation into the other equation.\u00a0The goal of the substitution method is to rewrite one of the equations in terms of a single variable. Then that equation can be solved for one of the variables. The second variable is found by substituting its new found value into either of the original equations and solving for the second variable.\r\n\r\nLet\u2019s start with an example to see what this means.\r\n
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      Example<\/h3>\r\nFind the solution of the system:\r\n\r\n[latex]4x+3y=\u221214\/\/y=2[\/latex]\r\n

      Solution<\/h4>\r\nThe second equation already gives us the value of\u00a0[latex]y[\/latex]:\u00a0\u00a0[latex]y=2[\/latex]\r\n\r\nWe are going to substitute this value of\u00a0\u00a0[latex]y=2[\/latex] into the first equation:\r\n

      [latex]4x+3y=\u221214[\/latex]<\/p>\r\n

      [latex]4x+3\\left(\\color{red}{2}\\right)=\u221214[\/latex]<\/p>\r\nNo simplify and solve the equation for [latex]x[\/latex].\r\n

      [latex]\\begin{array}{r}4x+6=\u221214\\\\4x=\u221220\\\\x=\u22125\\,\\,\\,\\end{array}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex](-5,2)[\/latex] is the solution of the system of equations.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWe can substitute a variable even if it is written as an expression.\r\n
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      Example<\/h3>\r\nSolve the system of equations:\r\n\r\n[latex]y+x=3\\\\x=y+5[\/latex]\r\n\r\n \r\n

      Solution<\/h4>\r\nThe goal of the substitution method is to rewrite one of the equations in terms of a single variable. The second equation tells us that [latex]x=y+5[\/latex], so we can substitute [latex]y+5[\/latex] for [latex]x[\/latex]\u00a0in the first equation:<\/span>\r\n

      [latex]\\begin{array}{r}y+\\color{red}{x}=3\\\\y+\\color{red}{\\left(y+5\\right)}=3\\end{array}[\/latex]<\/p>\r\nSimplify and solve the equation for [latex]y[\/latex]:\r\n

      [latex]\\begin{array}{r}2y+5=\\,\\,\\,\\,3\\\\\\underline{\u22125\\,\\,\\,\\,\\,\u22125}\\\\2y=\u22122\\\\y=\u22121\\end{array}[\/latex]<\/p>\r\nNow find [latex]x[\/latex] by\u00a0substituting this value for [latex]y[\/latex] into either equation and solving for [latex]x[\/latex].\r\n\r\nWe will use the first equation:\r\n

      [latex]\\begin{array}{r}\\color{red}{y}+x=3\\\\\\color{red}{\u22121}+x=3\\\\\\underline{+1\\,\\,\\,\\,\\,\\,\\,\\,\\,+1}\\\\x=4\\end{array}[\/latex]<\/p>\r\nFinally, check the solution\u00a0[latex]x=4[\/latex], [latex]y=\u22121[\/latex]\u00a0by substituting these values into each of the original equations.\r\n

      [latex]\\begin{array}{r}y+x=3\\\\\u22121+4=3\\\\3=3\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n

      [latex]\\begin{array}{l}x=y+5\\\\4=\u22121+5\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]x=4[\/latex] and [latex]y=\u22121[\/latex]\r\n\r\nThe solution is [latex](4,\u22121)[\/latex].\r\n\r\n<\/div>\r\nRemember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair [latex](4,\u22121)[\/latex] does work for both equations, so we know that it is a solution to the system as well.\r\n
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      Try It<\/h3>\r\nSolve the system of equations: [latex]2x+3y=6\\\\x=-3[\/latex]\r\n\r\n[reveal-answer q=\"hjm890\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm890\"]\r\n\r\n[latex](-3,4)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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      Try It<\/h3>\r\nSolve the system of equations: [latex]2x+3y=6\\\\y=3x+2[\/latex]\r\n\r\n[reveal-answer q=\"hjm808\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm808\"]\r\n\r\n[latex](0,2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the examples above, one of the equations was already given to us in terms of the variable [latex]x[\/latex] or [latex]y[\/latex]. This allowed us to quickly substitute that value into the other equation and solve for one of the unknowns.\r\n\r\nSometimes we have to rewrite one of the equations in terms of one of the variables first before we can substitute.\r\n
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      Example<\/h3>\r\nSolve the following system of equations by substitution.\r\n
      [latex]\\begin{equation}\\begin{aligned}-x+y & =-5\\\\ y & =x - 5\\ \\end{aligned}\\end{equation}[\/latex]<\/div>\r\n

      Solution<\/h4>\r\nSolve the first equation for [latex]y[\/latex]:\r\n
      [latex]\\begin{equation}\\begin{aligned}-x+y & =-5\\\\ y & =x - 5\\ \\end{aligned}\\end{equation}[\/latex]<\/div>\r\n
      <\/div>\r\nSubstitute the expression [latex]x - 5[\/latex] for [latex]y[\/latex] in the second equation:\r\n
      [latex]\\begin{equation}\\begin{aligned}2x - 5y & =1 \\\\ 2x - 5(x - 5)=1 \\\\ 2x - 5x+25 & =1\\\\ -3x & =-24 \\\\ x& =8\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n
      <\/div>\r\nSubstitute [latex]x=8[\/latex] into the first equation and solve for [latex]y[\/latex]:\r\n
      [latex]\\begin{equation}\\begin{aligned}-\\left(8\\right)+y & =-5 \\\\ y &=3\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n
      <\/div>\r\nOur solution is [latex]\\left(8,3\\right)[\/latex].\r\n\r\nCheck the solution by substituting [latex]\\left(8,3\\right)[\/latex] into both equations.\r\n

      [latex]\\begin{equation}\\begin{aligned}-x+y & =-5 \\\\ -(8)+(3) & =-5 \\;\\;\\;\\;\\;\\text{True}\\\\ \\\\2x - 5y & =1 \\\\ 2\\left(8\\right)-5\\left(3\\right) &=1 \\;\\;\\;\\;\\;\\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]\\left(8,3\\right)[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\nThe substitution method can be used to solve any linear system in two variables, but the method works best if one of the equations contains a coefficient of 1 or [latex]\u20131[\/latex] so that we do not have to deal with fractions.\r\n\r\nHere is a summary of the steps we use to solve systems of equations using the substitution method.\r\n
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      solvING A SYSTEM OF LINEAR EQUATIONS using the substitution method<\/h3>\r\n
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      1. Solve one of the two equations for one of the variables in terms of the other.<\/li>\r\n \t
      2. Substitute the expression for this variable into the unused equation, and then solve for the remaining variable.<\/li>\r\n \t
      3. Substitute that solution into either of the original equations to find the value of the other variable. If possible, write the solution as an ordered pair.<\/li>\r\n \t
      4. Check the solution in both equations.<\/li>\r\n<\/ol>\r\n<\/div>\r\nLet\u2019s look at some examples whose substitution involves the distributive property.\r\n
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        Example<\/h3>\r\nSolve the system of equations.\r\n

        [latex]\\begin{array}{l}y = 3x + 6\\\\\u22122x + 4y = 4\\end{array}[\/latex]<\/p>\r\n\r\n

        Solution<\/h4>\r\nThe first equation is already solved for [latex]y[\/latex] in terms of [latex]x[\/latex], so it makes sense to substitute [latex]3x + 6[\/latex] into the second equation for [latex]y[\/latex]:\r\n

        [latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122x+4\\left(3x+6\\right)=4\\end{array}[\/latex]<\/p>\r\nSimplify and solve the equation for [latex]x[\/latex]:\r\n

        [latex]\\begin{array}{r}\u22122x+12x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\10x+24=4\\,\\,\\,\\,\\,\\,\\,\\\\\\underline{\u221224\\,\\,\u221224\\,\\,\\,\\,}\\\\10x=\u221220\\\\x=\u22122\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nTo find[latex]y[\/latex], substitute this value for [latex]x[\/latex] back into one of the original equations.\r\n

        [latex]\\begin{array}{l}y=3x+6\\\\y=3\\left(\u22122\\right)+6\\\\y =\u22126+6\\\\y=0\\end{array}[\/latex]<\/p>\r\nCheck the solution [latex]x=\u22122[\/latex], [latex]y=0[\/latex]\u00a0by substituting them into each of the original equations.\r\n

        [latex]\\begin{array}{l}y=3x+6\\\\0=3\\left(\u22122\\right)+6\\\\0=\u22126+6\\\\0=0\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n

        [latex]\\begin{array}{r}\u22122x+4y=4\\\\\u22122\\left(\u22122\\right)+4\\left(0\\right)=4\\\\4+0=4\\\\4=4\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n

        Answer<\/h4>\r\n[latex]x=\u22122[\/latex] and [latex]y=0[\/latex]\r\n\r\nThe solution is [latex](\u22122, 0)[\/latex].\r\n\r\n<\/div>\r\n
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        Example<\/h3>\r\nSolve the system of equations.\r\n

        [latex]\\begin{array}{r}2x+3y=22\\\\3x+y=19\\end{array}[\/latex]<\/p>\r\n\r\n

        Solution<\/h4>\r\nThe second equation,[latex]3x+y=19[\/latex], can easily be rewritten in terms of [latex]y[\/latex]:\r\n

        [latex]\\begin{array}3x+y=19\\\\y=19\u20133x\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]19\u20133x[\/latex] for [latex]y[\/latex] in the other equation:\r\n

        [latex]\\begin{array}{r}2x+3y=22\\\\2x+3(19\u20133x)=22\\end{array}[\/latex]<\/p>\r\nSimplify and solve the equation for [latex]x[\/latex]:\r\n

        [latex]\\begin{array}{r}2x+57\u20139x=22\\,\\,\\,\\,\\\\\u22127x+57=22\\,\\,\\,\\,\\\\\u22127x=\u221235\\\\x=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nSubstitute [latex]x=5[\/latex] back into one of the original equations to solve for [latex]y[\/latex].\r\n

        [latex]\\begin{array}{r}3x+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\3\\left(5\\right)+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\15+y=19\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\\\y=19\u221215\\\\y=4\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nCheck both solutions by substituting them into each of the original equations.\r\n

        [latex]\\begin{array}{r}2x+3y=22\\\\2(5)+3\\left(4\\right)=22\\\\10+12=22\\\\22=22\\\\\\text{TRUE}\\\\\\\\3x+y=19\\\\3\\left(5\\right)+4= 19\\\\19=19\\\\\\text{TRUE}\\end{array}[\/latex]<\/p>\r\n\r\n

        Answer<\/h4>\r\n[latex]x=5[\/latex] and [latex]y=4[\/latex]\r\n\r\nThe solution is [latex](5, 4)[\/latex].\r\n\r\n<\/div>\r\nIf we had chosen the other equation to start with in the previous examples, we would still be able to find the same solution. \u00a0It is really a matter of preference because sometimes solving for a variable will result in having to work with fractions. \u00a0Gaining experience with algebra, will help us to anticipate what choices will lead to more desirable outcomes.\r\n
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        Try It<\/h3>\r\nSolve the system of equations.\r\n

        [latex]\\begin{array}{r}x+3y=6\\\\3y=x-6\\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"hjm460\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm460\"][latex](6,0)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video shows an example of solving a systems of two equations using the substitution method.\r\n\r\n