3y-8(-19+4y)=65

Question

Actually, we're asked to Solve the system of equations using substitution method for the following.... 

x-4y=-19 and 3y-8x=65

Answer 1

Start by solving the first equation for x.

x - 4y = -19

x = 4y - 19

Now plug this in the second equation.

3y - 8x = 65

3y - 8(4y - 19) = 65

3y - 32y + 152 = 65

-29y = 65 - 152 = -87

y = -87 / -29 = 3

Plug this in the first equation.

x = 4y - 19 = 4*3 - 19 = -7

So x = -7, y = 3 is the solution.

We can check this answer:

x - 4y = -19

-7 - 4(3) = -19

-7 - 12 = -19

-19 = -19

and

3y - 8x = 65

3(3) - 8(-7) = 65

9 + 56 = 65

65 = 65

 

Answer 2

Maranda,

To answer systems of equaltions by substitution first solve one varible in terms of the other then substitute in for that varible.  Here solve x-4y = -19 for x.  Add 4y to each side of the equation to have x = -19 +4y.  Substitute this in to the original equation which I assume was 3y-8x = 65.  Now you have 3y-8(-19+4y) = 65.

Multiply through to have 3y-162+32y =65.  Add 162 to each side and combine like terms.  -29y=-97.  Divide through and find y = 3.  Substitute in for y into the equation that is solved for x  (x=-19+4y). 

Now x = -19+4*3.  Solve and find x =-19 +12 or x = -7.

Rebecca Muegge

Wilton, CA