Solving with Elimination Practice

Question

Complete this assessment to review what you've learned. It will not count toward your grade.
4 of 54 of 5 Items

Question
Solve the system of equations.
–x+6y=13

4x–10y=–10
(1 point)
(
,
)

Answer 1

To solve the system of equations using elimination, we start with the two equations:

1. \(-x + 6y = 13\) (Equation 1)
2. \(4x - 10y = -10\) (Equation 2)

First, we can multiply Equation 1 by 4 to align the coefficients of \(x\):

\[ 4(-x + 6y) = 4(13) \] \[ -4x + 24y = 52 \quad \text{(Equation 3)} \]

Now our system of equations is:

1. \(-4x + 24y = 52\) (Equation 3)
2. \(4x - 10y = -10\) (Equation 2)

Next, we add Equation 3 and Equation 2:

\[ (-4x + 24y) + (4x - 10y) = 52 - 10 \] \[ (24y - 10y) = 42 \] \[ 14y = 42 \] \[ y = 3 \]

Now that we have \(y\), we substitute \(y = 3\) back into one of the original equations to find \(x\). We can use Equation 1:

\[ -x + 6(3) = 13 \] \[ -x + 18 = 13 \] \[ -x = 13 - 18 \] \[ -x = -5 \] \[ x = 5 \]

So the solution to the system of equations is:

\[ (x, y) = (5, 3) \]

Final Answer: \((5, 3)\)