Actually, we're asked to Solve the system of equations using substitution method for the following....
x-4y=-19 and 3y-8x=65
Actually, we're asked to Solve the system of equations using substitution method for the following....
x-4y=-19 and 3y-8x=65
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
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