Solving with Elimination Practice

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

4x–10y=–10
(1 point)
(
,
)

1 answer

To solve the system of equations using the elimination method, we can start with the given equations:

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

First, let's manipulate the equations to eliminate one of the variables. We can multiply Equation 1 by 4 in order to align the coefficients of \(x\) for elimination:

\[ 4(-x + 6y) = 4(13) \]

This gives us:

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

Now we have:

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

Next, we can add Equation 2 and Equation 3 together to eliminate \(x\):

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

This simplifies to:

\[ (24y - 10y) = 42 \]

So we have:

\[ 14y = 42 \]

Now, divide both sides by 14:

\[ y = \frac{42}{14} = 3 \]

Now that we have the value of \(y\), we can substitute it back into one of the original equations to find \(x\). Let's use Equation 1:

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

This simplifies to:

\[ -x + 18 = 13 \]

Now, isolate \(x\):

\[ -x = 13 - 18 \] \[ -x = -5 \]

Thus:

\[ x = 5 \]

So the solution to the system of equations is:

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

In coordinate form, the solution is:

(5, 3).