To solve the given system of linear equations:
[tex]\[
\begin{cases}
3x - 4y = -6 \\
2x + 4y = 16
\end{cases}
\][/tex]
we will use the method of elimination.
1. Add the equations to eliminate [tex]\(y\)[/tex]:
Let's add the two equations together. This will allow us to eliminate the [tex]\(y\)[/tex] term since the coefficients of [tex]\(y\)[/tex] are opposites ([tex]\(-4y\)[/tex] and [tex]\(+4y\)[/tex]).
[tex]\[
(3x - 4y) + (2x + 4y) = -6 + 16
\][/tex]
Simplify:
[tex]\[
3x + 2x - 4y + 4y = -6 + 16
\][/tex]
[tex]\[
5x = 10
\][/tex]
2. Solve for [tex]\(x\)[/tex]:
[tex]\[
5x = 10
\][/tex]
Divide both sides by 5:
[tex]\[
x = 2
\][/tex]
3. Substitute [tex]\(x = 2\)[/tex] back into one of the original equations to solve for [tex]\(y\)[/tex]:
We can use either of the original equations. Let's use the first one:
[tex]\[
3x - 4y = -6
\][/tex]
Substitute [tex]\(x = 2\)[/tex]:
[tex]\[
3(2) - 4y = -6
\][/tex]
Simplify:
[tex]\[
6 - 4y = -6
\][/tex]
Subtract 6 from both sides:
[tex]\[
-4y = -6 - 6
\][/tex]
[tex]\[
-4y = -12
\][/tex]
Divide both sides by -4:
[tex]\[
y = 3
\][/tex]
4. Conclusion:
The solution to the system of equations is:
[tex]\[
x = 2, \quad y = 3
\][/tex]
So, the point of intersection of the two lines described by the system of equations is [tex]\((2, 3)\)[/tex].
