Solving with Elimination Practice

Question

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Question
Solve the system of equations.
3x+6y=21

–5x+2y=13
(1 point)
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,
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Answer 1

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

1. \( 3x + 6y = 21 \) (Equation 1)
2. \( -5x + 2y = 13 \) (Equation 2)

Step 1: Make the coefficients of y the same.

To do this, we can multiply Equation 2 by 3 to make the coefficient of y in both equations equal:

1. \( 3x + 6y = 21 \) (remains the same)
2. \( 3(-5x + 2y) = 3(13) \)
   \( -15x + 6y = 39 \) (Equation 3)

Now we have the following system of equations:

1. \( 3x + 6y = 21 \)
2. \( -15x + 6y = 39 \)

Step 2: Eliminate y.

Now we will subtract Equation 1 from Equation 3:

\[ (-15x + 6y) - (3x + 6y) = 39 - 21 \]

This simplifies to:

\[ -15x - 3x + 6y - 6y = 18 \] \[ -18x = 18 \]

Step 3: Solve for x.

Divide both sides by -18:

\[ x = -1 \]

Step 4: Substitute x back to find y.

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

\[ 3(-1) + 6y = 21 \] \[ -3 + 6y = 21 \] \[ 6y = 21 + 3 \] \[ 6y = 24 \] \[ y = 4 \]

Step 5: Write the solution.

The solution to the system of equations is:

\[ (x, y) = (-1, 4) \]

So the final answer is:

\[ \boxed{(-1, 4)} \]