What is the solution to the system of equations?

[tex]\[
\left\{
\begin{array}{l}
-2x + 5y = -16 \\
3x - 4y = 17
\end{array}
\right.
\][/tex]

A. [tex]\(\left( 8 \frac{1}{3}, 2 \right)\)[/tex]

B. [tex]\(\left( -3, -4 \frac{2}{5} \right)\)[/tex]

C. [tex]\((-2, 3)\)[/tex]

D. [tex]\((3, -2)\)[/tex]



Answer :

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]