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]
[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]