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 \((-1, 4)\).
Solve the system of equations.
3x+6y=21
–5x+2y=13 (1 point)
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WITH THE CORRECT AND SIMPLIFIED ANSWERS
1 answer