To solve the system of equations:
[tex]\[
\left\{\begin{array}{l}
-2x + 5y = -16 \\
3x - 4y = 17
\end{array}\right.
\][/tex]
we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. Let's go through the steps methodically.
1. First Equation:
[tex]\[
-2x + 5y = -16
\][/tex]
2. Second Equation:
[tex]\[
3x - 4y = 17
\][/tex]
Multiply the first equation by 3 and the second equation by 2 to make the coefficients of [tex]\( x \)[/tex] the same:
[tex]\[
3(-2x + 5y) = 3(-16) \implies -6x + 15y = -48
\][/tex]
[tex]\[
2(3x - 4y) = 2(17) \implies 6x - 8y = 34
\][/tex]
3. Resulting System:
[tex]\[
\left\{
\begin{array}{l}
-6x + 15y = -48 \\
6x - 8y = 34
\end{array}
\right.
\][/tex]
Add the two equations together to eliminate [tex]\( x \)[/tex]:
[tex]\[
(-6x + 15y) + (6x - 8y) = -48 + 34
\][/tex]
[tex]\[
7y = -14
\][/tex]
Solve for [tex]\( y \)[/tex]:
[tex]\[
y = -2
\][/tex]
4. Substitute [tex]\( y \)[/tex] back into the first equation to solve for [tex]\( x \)[/tex]:
[tex]\[
-2x + 5(-2) = -16
\][/tex]
[tex]\[
-2x - 10 = -16
\][/tex]
[tex]\[
-2x = -6
\][/tex]
[tex]\[
x = 3
\][/tex]
Thus, the solution to the system of equations is [tex]\( x = 3 \)[/tex] and [tex]\( y = -2 \)[/tex].
To verify, we substitute [tex]\( x = 3 \)[/tex] and [tex]\( y = -2 \)[/tex] back into both original equations:
First Equation:
[tex]\[
-2(3) + 5(-2) = -6 - 10 = -16
\][/tex]
Second Equation:
[tex]\[
3(3) - 4(-2) = 9 + 8 = 17
\][/tex]
Both equations are satisfied, so the solution is correct.
Therefore, the solution to the system of equations is:
[tex]\[
(3, -2)
\][/tex]
