Question

How do I solve each system by elimination

`6x + 8y = - 10 \\ y = - 5x + 3`

Answer 1

The system can be solved to form a solution of (1,-2). First, you have to convert y = -5x+3 to standard form. Then, you need to multiply the 1y (from the second equation) by -8 so that they “y”’s cancel out (or eliminate). Then, add both of the equations together and solve for x. After that, plug in the x value that you got into one of the original equations — it doesn’t matter which one — and solve for y. Tada! You have your answer, (1,-2). I hope this helped!! :))

Answer 2

Solve the following system by elimination:
{6 x + 8 y = -10 | (equation 1)
{y = 3 - 5 x | (equation 2)

Express the system in standard form:
{6 x + 8 y = -10 | (equation 1)
{5 x + y = 3 | (equation 2)


Subtract 5/6 × (equation 1) from equation 2:
{6 x + 8 y = -10 | (equation 1)
{0 x - (17 y)/3 = 34/3 | (equation 2)


Divide equation 1 by 2:
{3 x + 4 y = -5 | (equation 1)
{0 x - (17 y)/3 = 34/3 | (equation 2)

Multiply equation 2 by 3/17:
{3 x + 4 y = -5 | (equation 1)
{0 x - y = 2 | (equation 2)

Multiply equation 2 by -1:
{3 x + 4 y = -5 | (equation 1)
{0 x+y = -2 | (equation 2)

Subtract 4 × (equation 2) from equation 1:
{3 x+0 y = 3 | (equation 1)
{0 x+y = -2 | (equation 2)

Divide equation 1 by 3:
{x+0 y = 1 | (equation 1)
{0 x+y = -2 | (equation 2)


Collect results:

{Answer: {x = 1, y = -2