Solving Systems of Equations by Addition / Elimination

October 6, 2010 at 1:32 pm 2 comments

Many algebra students struggle at first when using the addition method (aka the elimination method) to solve a system of two linear equations in two unknowns. The hardest parts seem to be deciding whether to multiply one equation or two, and exactly what constant to multiply by the equation(s). In this blog I’ll go over strategies to make you an effective “systems solver”.

Goal: Once the equations are written in standard form (Ax+By=C), the goal is to make either the coefficients of the x terms or the coefficients of the y terms opposites of each other. This will allow you to add the two equations together, leaving you with one equation with only one unknown.

Tools: You can multiply both sides of either equation by a constant. You can do this for one equation or both.

Case 1: The Dream Scenario

The easiest application of the addition method occurs when the coefficients of the x terms or the y terms are already opposites.

\begin{array}{rcl} 3x-2y & = & 13 \\ 7x+2y & = & 37 \end{array}

Add the equations to get 10x=50, which can be solved by dividing both sides of the equation by 10. The solution is x=5.

Now you can substitute 5 for x into either of the two original equations to solve for y. 7(5)+2y=37 leads to y=1. The solution of the system is (5,1).

Case 2: One Coefficient is a Multiple of the Other

If the coefficient of the x terms or the coefficients of the y terms are not already multiples, check to see if one coefficient of a term containing x is a multiple of the other term containing x. (If not, also check the terms containing y for the same.) If one coefficient is a multiple of the other, then you can solve the system by multiplying only one equation on both sides by a constant.

\begin{array}{rcl} 4x+5y & = & 39 \\ 2x+3y & = & 21 \end{array}

Look at the terms containing x and notice that 4 is a multiple of 2. It will be easy to change 2x into -4x by multiplying both sides of the second equation by -2. (We need -4x, not 4x, so that the two coefficients are opposites.)

\begin{array}{rcl} 4x+5y & = & 39 \\ -2(2x+3y) & = & -2(21) \end{array}

\begin{array}{rcl} 4x+5y & = & 39 \\ -4x-6y & = & -42 \end{array}

When you add, the resulting equation is -y=-3. Once you solve that equation, you’ll find that y=3.

Back substituting to find x:

4x+5y=39 \\ 4x+5(3)=39 \\ 4x+15=39 \\ 4x=24 \\ x=6

The solution is (6,3).

Case 3: Multiplying Both Equations

If the neither pair of coefficients are multiples, you must multiply both equations by a constant. (The constant for the first equation will be different than the second equation.)

\begin{array}{rcl} 6x+3y & = & 30 \\ 8x-4y & = & 56 \end{array}

The coefficients of the terms containing x are not multiples. The coefficients of the terms containing y are not multiples either. In this case you have to multiply both equations by a constant.

So, which variable should we seek to eliminate? Because the terms containing y have one positive coefficient and one negative coefficient, seeking to eliminate y is a good choice because you won’t have to multiply by a negative number. The coefficients are 3 and -4. What is the LCM of 3 and 4? It’s 12, and the goal is to multiply each equation in such a way that 12y or -12y results. This can be done by multiplying the first equation by 3 and the second equation by 4.

\begin{array}{rcl} 4(6x+3y) & = & 4(30) \\ 3(8x-4y) & = & 3(56) \end{array}

This produces the following system.

\begin{array}{rcl} 24x+12y & = & 120 \\ 24x-12y & = & 168 \end{array}

When you add, the resulting equation is 48x=288. Once you solve that equation, you’ll find that x=6.

Back substituting to find x:

6x+3y=30 \\ 6(6)+3y=30 \\ 36+3y=30 \\ 3y=-6 \\ y=-2

The solution is (6,-2).

Summary

I hope that you find this helpful. If you have any questions, or if you have any other topics you are interested in, you can reach me through the contact page at my web site – georgewoodbury.com.

-George

I am a math instructor at College of the Sequoias in Visalia, CA. Each Wednesday I post an article related to General Teaching on my blog. If there’s a particular topic you’d like me to address, or if you have a question or a comment, please let me know. You can reach me through the contact page on my website – http://georgewoodbury.com.

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Entry filed under: General Teaching, Math. Tags: , , , , , , , , , , , , , , , , .

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2 Comments Add your own

  • 1. kent  |  October 13, 2010 at 4:33 am

    thank you for your help i wish you help many more peolpe

    Reply
  • 2. kent  |  October 13, 2010 at 4:33 am

    sorry wrong spelling

    Reply

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