To solve the system of equations using elimination, we start with the given system:
[tex]\[
\text{Equation 1:} \quad -5x - y = 9
\][/tex]
[tex]\[
\text{Equation 2:} \quad 3x - 5y = 17
\][/tex]
Step 1: Eliminate [tex]\( y \)[/tex]
We want to find a way to eliminate one of the variables by combining the two equations. To eliminate [tex]\( y \)[/tex], we first manipulate the equations so that the coefficients of [tex]\( y \)[/tex] are the same (or opposites) in both equations.
Let's multiply Equation 1 by [tex]\( 5 \)[/tex] to make the coefficient of [tex]\( y \)[/tex] equal to [tex]\( -5 \)[/tex] (the same as in Equation 2).
[tex]\[
5 \times (-5x - y) = 5 \times 9
\][/tex]
[tex]\[
-25x - 5y = 45
\][/tex]
Now we have a new system:
[tex]\[
\text{Equation 3:} \quad -25x - 5y = 45 \quad \text{(modified Equation 1)}
\][/tex]
[tex]\[
\text{Equation 2:} \quad 3x - 5y = 17
\][/tex]
Step 2: Subtract the modified Equation 1 from Equation 2
Now, subtract Equation 2 from the modified Equation 1:
[tex]\[
(-25x - 5y) - (3x - 5y) = 45 - 17
\][/tex]
[tex]\[
-25x - 3x = 45 - 17
\][/tex]
[tex]\[
-28x = 28
\][/tex]
Step 3: Solve for [tex]\( x \)[/tex]
Divide both sides of the equation by [tex]\(-28\)[/tex]:
[tex]\[
x = \frac{28}{-28}
\][/tex]
[tex]\[
x = -1
\][/tex]
Step 4: Substitute [tex]\( x \)[/tex] back to find [tex]\( y \)[/tex]
Now that we have [tex]\( x = -1 \)[/tex], substitute [tex]\( x \)[/tex] into one of the original equations to solve for [tex]\( y \)[/tex]. We use Equation 1:
[tex]\[
-5(-1) - y = 9
\][/tex]
[tex]\[
5 - y = 9
\][/tex]
[tex]\[
-y = 9 - 5
\][/tex]
[tex]\[
-y = 4
\][/tex]
[tex]\[
y = -4
\][/tex]
Solution
The solution to the system of equations is:
[tex]\[
x = -1 \quad \text{and} \quad y = -4
\][/tex]
Therefore, the solution is:
[tex]\[
(x, y) = (-1, -4)
\][/tex]
