# How do you determine whether a linear system has one solution, many solutions, or no solution when given 8x =+ y = -16 and -3x + y = - 5?

Apr 10, 2018

There is one solution. $x = - 1$ and $y = - 8$

#### Explanation:

I'm assuming the two equations are

Equation I: $8 x + y = - 16$, and

Equation II: $- 3 x + y = - 5$.

I will subtract Equation II from Equation I.

$8 x + y - \left(- 3 x\right) - y = - 16 - \left(- 5\right)$

Which simplifies to

$11 x = - 11$.

Divide both sides by 11.

$x = - 1$

Substitute $x = - 1$ into Equation II.

$- 3 \left(- 1\right) + y = - 5$.

Simplify.

$3 + y = - 5$

Subtract three from both sides of this equation.

$y = - 8$

There is one solution. $x = - 1$ and $y = - 8$