Question

solve the following system of equations using elimination -6x+10y=-16 6x-3y=12. -6+10y=-16, 6x-3y=-12/ y=___6x-3(__)=-12 6x-__=-12 6x=__ y=___ the solution is (___,___)
the following answers that can be used in this are: 12,-28,4,24,-4,-24,4/7,-4/7,0,-12,28,7

Answer 1

To solve the given system of equations using elimination, we have:

1) \(-6x + 10y = -16\)
2) \(6x - 3y = 12\)

First, we can add the two equations together to eliminate \(x\):

\[
-6x + 10y + 6x - 3y = -16 + 12
\]

This simplifies to:

\[
(10y - 3y) = -16 + 12
\]

\[
7y = -4
\]

Now, solve for \(y\):

\[
y = \frac{-4}{7}
\]

Now that we have \(y\), we can substitute this value back into one of the original equations to solve for \(x\). We'll use the second equation:

\[
6x - 3y = 12
\]

Substituting \(y = \frac{-4}{7}\):

\[
6x - 3\left(\frac{-4}{7}\right) = 12
\]

\[
6x + \frac{12}{7} = 12
\]

Now, subtract \(\frac{12}{7}\) from both sides:

\[
6x = 12 - \frac{12}{7}
\]

To perform this subtraction, convert \(12\) into a fraction with a common denominator:

\[
12 = \frac{84}{7}
\]

Now, we can substitute this back in:

\[
6x = \frac{84}{7} - \frac{12}{7}
\]

\[
6x = \frac{84 - 12}{7}
\]

\[
6x = \frac{72}{7}
\]

Now, divide by \(6\) to solve for \(x\):

\[
x = \frac{72}{7} \cdot \frac{1}{6} = \frac{72}{42} = \frac{12}{7}
\]

So, we have:

\[
x = \frac{12}{7} \quad \text{and} \quad y = -\frac{4}{7}
\]

Finally, the solution to the system of equations is:

\[
\left(\frac{12}{7}, -\frac{4}{7}\right)
\]

In terms of the options provided:

\[
\text{The solution is } \left(12/7, -4/7\right)
\]