Answer :
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]
[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]
