Question

If you were to use the substitution method to solve the following system, choose the new system of equations that would result if x was isolated in the second equation.

5x – 2y – z = 16
x + y + 4z = 6
3x + 2y + 6z = 16

Choices:



A)–7y – 21z = –14
–y – 6z = –2

B)7x + 7z = 28
x – 2z = 4

C)–3y + 19z = –14
5y + 18z = –2

D)21x – 7y = 70
33x – 10y = 112

Answer 1

The answer is A.

if you isolate x in eq2, you get x = (6 - y - 4z)...

plug that into eq1 and you get...

5(6 - y - 4z) -2y -z = 16...

30 -5y -20z -2y -z = 16...

-7y - 21z = 16-30 = -14...

-7y - 21z = -14...

Now plugging (6 - y - 4z) into eq3...

3(6 - y - 4z) +2y +6z = 16
18 - 3y -12z +2y +6z = 16
18 -1y -6z = 16
-y-6z = 16 - 18 = -2
-y-6z = -2

Answer 2

A.
All you have to do is R1 - 5xR2 and R3 - 3xR2. R1, R2, and R3 are equation 1, 2, and 3 respectively. You multiply R2 by 5 in the first because R1 has a 5x, and R2 only has x. The same reason applies for the second subtraction.

Answer 3

ok thank u :)

Answer 4

Hmmm. Is that the same as the substitution method?

Answer 5

Good point, I didn't read the problem carefully enough. You can still stop once you reach -7y - 21z = -14 though.