Question

solve the system algebraically using either the substitution or elimination method. if there is exactly one solution, write as an ordered pair. if not, choose one of the other options.
\\(\

$$\begin{cases}-6x + 5y = 8 \\ -5x + 20y = -25\\end{cases}$$

\\)
o one or more solutions:
o no solution
o infinite number of solutions

Explanation:

Step1: Multiply first - equation for elimination

Multiply the first equation $-6x + 5y=8$ by $4$ to make the coefficient of $y$ the same as in the second equation.
$4(-6x + 5y)=4\times8$, which gives $-24x+20y = 32$.

Step2: Subtract the second equation from the new - formed equation

Subtract the second equation $-5x + 20y=-25$ from $-24x+20y = 32$.
$(-24x+20y)-(-5x + 20y)=32-(-25)$.
$-24x + 20y + 5x-20y=32 + 25$.
$-19x=57$.

Step3: Solve for $x$

Divide both sides of $-19x = 57$ by $-19$.
$x=\frac{57}{-19}=-3$.

Step4: Substitute $x$ into the first equation to solve for $y$

Substitute $x = - 3$ into $-6x+5y = 8$.
$-6\times(-3)+5y=8$.
$18 + 5y=8$.
$5y=8 - 18=-10$.
$y=\frac{-10}{5}=-2$.

Answer:

One or more solutions: $(-3,-2)$