Solve the system of equations.

3x+6y=21
–5x+2y=13 (1 point)
(
,
)
WITH THE CORRECT AND SIMPLIFIED ANSWERS

1 answer

Let's solve the system of linear equations using the method of substitution or elimination. In this case, using elimination seems straightforward since we can eliminate y by multiplying the second equation so that the coefficients of y in both equations are opposites.

First, the system of equations is:

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

To eliminate y, we will multiply the second equation by 3 so that the coefficient of y in the second equation will be 6, the same as in the first equation but with opposite sign.

Multiplying the second equation by 3:

\(3(-5x + 2y) = 3(13)\)

Which gives us:

\(-15x + 6y = 39\)

Now, the system is:

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

Subtracting equation 1) from 3):

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

\(-15x + 6y - 3x - 6y = 18\)

\(-18x = 18\)

Solving for x:

\(x = \frac{18}{-18}\)

\(x = -1\)

Now that we have x, we can substitute it into one of the original equations to get y. We can use the first equation:

\(3x + 6y = 21\)

Substitute x with -1:

\(3(-1) + 6y = 21\)

\(-3 + 6y = 21\)

Add 3 to both sides:

\(6y = 24\)

Divide by 6:

\(y = \frac{24}{6}\)

\(y = 4\)

Therefore, the solution to the system of equations is \(x = -1\) and \(y = 4\), so the point of intersection where both lines meet is \((-1, 4)\).