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How do I solve systems by substitution?

4x+2y=-2

y=6x-5

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Tamara J. | Math Tutoring - Algebra and Calculus (all levels)Math Tutoring - Algebra and Calculus (al...
4.9 4.9 (51 lesson ratings) (51)
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The system consists of the following equations:     4x + 2y = -2     and     y = 6x - 5

To solve for the system by the substitution method, we take one of the equations and solve it for either variable (x= or y=) then substitute this into the other equation. Here, we see that the second equation has already been solved for one of the variables, that variable being y. So we substitute what y is equal to in terms of x into the first equation, thus being able to solve for x. Once we've solved for x, we use its value to solve for y.

Since   y = 6x - 5 , then

     4x + 2y = -2                        substitute 6x - 5 for y

     4x + 2(6x - 5) = -2              multiply each term inside the parentheses by 2

     4x + 2·6x + 2·-5 = -2

     4x + 12x - 10 = -2                combine like terms on the left hand side of the equation

     16x - 10 = -2                        add 10 to both sides of the equation

     16x - 10 + 10 = -2 + 10

     16x = 8                                divide both sides of the equation by 16 to solve for x

     16x/16 = 8/16

     x = 1/2

To solve for y, plug the value of x into either of the two original equations:

     y = 6x - 5   ,     x = 1/2

     y = 6(1/2) - 5

     y = 3 - 5

     y = -2

Kurt T. | Math Tutoring and Test PrepMath Tutoring and Test Prep
4.9 4.9 (115 lesson ratings) (115)
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In this case, you could substitute (6x-5) for y in the first equation, solve for x, and substitute the new x value into both equations to make sure you get the same answer for y...

4x + 2 (6x - 5) = -2

4x + 12x - 10 = -2

16x = 8

x = 1/2

Now substitute 1/2 for x in both equations...

4 (1/2) + 2y = -2

2 + 2y = -2

2y = -4

y = -2

y = 6 (1/2) - 5

y = 3 - 5

y = -2