{"id":958,"date":"2021-10-08T19:23:35","date_gmt":"2021-10-08T19:23:35","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&p=958"},"modified":"2022-03-10T23:18:29","modified_gmt":"2022-03-10T23:18:29","slug":"7-3-1-the-elimination-method-without-multiplication","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/7-3-1-the-elimination-method-without-multiplication\/","title":{"raw":"7.3.1: The Elimination Method I","rendered":"7.3.1: The Elimination Method I"},"content":{"raw":"
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Learning Outcomes<\/h3>\r\n
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  • Use the elimination method without multiplication<\/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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    • addition property of equality<\/strong>: adding the same term to both sides of an equation results in an equivalent equation<\/li>\r\n \t
    • elimination method:\u00a0<\/strong> a method that uses the addition property of equality to solve a system of linear equations<\/li>\r\n \t
    • opposites<\/strong>: two terms that are identical in absolute value but opposite in sign<\/li>\r\n \t
    • coefficient<\/strong>: the number multiplied onto a variable<\/li>\r\n<\/ul>\r\n<\/div>\r\n

      The Elimination Method<\/h2>\r\nThe\u00a0elimination method<\/em>\u00a0<\/b>for solving systems of linear equations uses the addition property of equality:\u00a0<\/strong><\/em>adding the same term to both sides of an equation results in an equivalent equation. This can be extended to: adding two equations together results in an equivalent equation. This is true because if [latex]A=B[\/latex] and [latex]C=D[\/latex], then adding [latex]C[\/latex] to both sides of the first equation is equivalent to adding [latex]D[\/latex] to both sides of the first equation, since\u00a0[latex]C=D[\/latex]. That means that\u00a0[latex]A+C=B+C[\/latex] and [latex]A+D=B+D[\/latex]. Consequently,\u00a0[latex]A+C=B+D[\/latex] is also an equivalent equation since [latex]C=D[\/latex].\r\n\r\nFor example, if [latex]x+y=3\\\\x-y=5[\/latex] then adding the equations together gives [latex]2x=8[\/latex], which is an equivalent equation to both [latex]x+y=3[\/latex] and\u00a0[latex]x-y=5[\/latex]. In addition, adding the two equations together\u00a0eliminated<\/em> the [latex]y[\/latex]-terms since they were opposites<\/strong><\/em>, and their sum is zero. This results in a linear equation in one variable\u00a0[latex]2x=8[\/latex], which can be solved to\u00a0[latex]x=4[\/latex]. Now that we know the value of the\u00a0[latex]x[\/latex]-variable, we can substitute its value into either of the original equations to find\u00a0[latex]y[\/latex]:\u00a0[latex]x+y=3\\\\4+y=3\\\\y=-1[\/latex] Consequently the solution of the system is [latex](4,-1)[\/latex].\r\n\r\nThis solution must also satisfy the other equation\u00a0[latex]x-y=5[\/latex], which it does:\u00a0[latex]4-(-1)=5[\/latex].\r\n\r\nIn order for the elimination method to work, one of the variables must be eliminated when the equations are added together. This requires that the coefficients<\/strong><\/em> of one of the variables are opposites.\r\n
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      Example<\/h3>\r\nUse elimination to solve the system.\r\n

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

      Solution<\/h4>\r\nThe [latex]y[\/latex] terms have opposite coefficients of [latex]-1[\/latex] and\u00a0[latex]1[\/latex], which add to zero.\r\n\r\nAdd the equations:\r\n

      [latex]\\begin{equation}\\begin{aligned}x-y &=-6\\\\ x+y & =8\\\\ \\\\2x +0y& =2\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nSolve for [latex]x[\/latex]:\r\n

      [latex]\\begin{equation}\\begin{aligned}2x &=2\\\\x &=1\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nSubstitute [latex]x=1[\/latex] into one of the original equations and solve for [latex]y[\/latex]:\r\n

      [latex]\\begin{equation}\\begin{aligned}x+y &=8\\\\1+y &=8\\\\y&=8\u20131\\\\y&=7\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nSince we used the second equation to find [latex]x[\/latex], we can check the solution in the first equation:\r\n

      [latex]\\begin{equation}\\begin{aligned}x\u2013y&=\u22126\\\\1\u20137&=\u22126\\\\\u22126&=\u22126\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nThe answer checks.\r\n

      Answer<\/h4>\r\nThe solution to this system is [latex]\\left(1,7\\right)[\/latex].\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
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      Example<\/h3>\r\nSolve the given system of equations by elimination.\r\n
      [latex]\\begin{equation}\\begin{aligned}x+2y&=-1\\\\ -x+y&=3\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n

      Solution<\/h4>\r\nThe [latex]x[\/latex] terms have opposite coefficients of [latex]1[\/latex] and\u00a0[latex]-1[\/latex], which add to zero.\r\n\r\nAdd the two equations together to eliminate [latex]x[\/latex]:\r\n
      [latex]\\begin{equation}\\begin{aligned}x+2y&=-1 \\\\ -x+y&=3 \\\\ \\\\0x+3y&=2\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n \r\n\r\nNow that we have eliminated [latex]x[\/latex], we can solve the resulting equation for [latex]y[\/latex]:\r\n

      [latex]\\begin{equation}\\begin{aligned}3y&=2 \\\\ y&=\\frac{2}{3}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nThen, we substitute this value for [latex]y[\/latex] into one of the original equations and solve for [latex]x[\/latex]:\r\n

      [latex]\\begin{equation}\\begin{aligned}-x+y&=3 \\\\ -x+\\frac{2}{3}&=3\\\\-x&=3-\\dfrac{2}{3} \\\\ -x&=\\frac{7}{3} \\\\ x&=-\\frac{7}{3}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nThe solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].\r\n\r\nSince we used the second equation to find [latex]x[\/latex], we can check the solution in the first equation:\r\n

      [latex]\\begin{equation}\\begin{aligned}x+2y&=-1\\\\ \\left(-\\dfrac{7}{3}\\right)+2\\left(\\dfrac{2}{3}\\right)&=-1 \\\\ -\\dfrac{7}{3}+\\dfrac{4}{3}&=-1\\\\ \\text{ }-\\dfrac{3}{3}&=-1\\\\ -1&=-1\\;\\;\\;\\;\\; \\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\nThe solution to this system is [latex]\\left(-\\dfrac{7}{3},\\dfrac{2}{3}\\right)[\/latex].\r\n\r\n<\/div>\r\nWe gain an important perspective on systems of equations by looking at the graphical representation. In the graph below, we see that the equations intersect at the solution. We do not need to ask whether there may be a second solution, because observing the graph confirms that the system has exactly one solution.\r\n\r\n\"A\r\n
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      Try It<\/h3>\r\nSolve the given system of equations by elimination.\r\n
      [latex]\\begin{equation}\\begin{aligned}x+3y&=6\\\\ -x+3y&=-6\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n
      [reveal-answer q=\"hjm801\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm801\"][latex](0,6)[\/latex][\/hidden-answer]<\/div>\r\n
      <\/div>\r\n<\/div>\r\n
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      Try It<\/h3>\r\nSolve the given system of equations by elimination.\r\n
      [latex]\\begin{equation}\\begin{aligned}x+y&=4\\\\ x-y&=-6\\end{aligned}\\end{equation}[\/latex]<\/div>\r\n
      [reveal-answer q=\"hjm354\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm354\"][latex](-1,5)[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n \r\n\r\nUnfortunately not all systems work out this nicely. How about a system like [latex]2x+y=12[\/latex] and [latex]\u22123x+y=2[\/latex]. If we add these two equations together, no variables are eliminated: [latex]-x+2y=14[\/latex]\r\n\r\nBut we need to eliminate a variable for this method to work. Each equation has an [\/latex]x[\/latex] variable with a coefficient of [latex]1[\/latex], so if we were to subtract the equations[\/latex]x[\/latex] would be eliminated. Subtraction is equivalent to adding the opposite, so we add the opposite of one of the equations to the other equation. This means multiplying every term in one of the equations by [latex]-1[\/latex]: [latex]-1\\left(-3x+y\\right )=2(-1)[\/latex] becomes [latex]3x-y=-2[\/latex].\r\n

      [latex]\\begin{equation}\\begin{aligned}2x+y&=12\\\\3x-y&=-2\\\\ 5x\\;\\;\\;\\;\\;&=10\\\\x&=2\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nWe eliminated the [latex]y[\/latex] variable, and solved for [latex]x[\/latex]. We can now substitute [latex]x[\/latex] into either of the original equations to solve for [latex]y[\/latex]:\r\n

      [latex]\\begin{equation}\\begin{aligned}2x+y&=12\\\\2(2)+y&=12\\\\4+y&=12\\\\y&=8\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nConsequently, the solution is [latex](2, 8)[\/latex], which we can check by substituting it into the second equation:\r\n

      [latex]\\begin{equation}\\begin{aligned}-3x+y&=2\\\\2(2)+8&=12\\\\4+8&=12\\\\12&=12\\;\\;\\;\\text{True}\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n \r\n\r\nAs long as the coefficients are opposites we can eliminate the variable through addition.\u00a0The following video describes a similar problem where one variable is eliminated by adding the two equations together.<\/span>\r\n\r\n