Solve the system of equations. -9x + 10y = 3 3x - 2y = -9 (____, ____)

Write Equations: Step 1: Write down the system of equations. -9x + 10y = 3 3x - 2y = -9Eliminate y: Step 2: Use elimination method to eliminate y. Multiply the second equation by 5 to balance the coefficients of y. 5(3x - 2y) = 5(-9) 15x - 10y = -45Add Equations: Step 3: Add the modified second equation to the first equation to eliminate y. (-9x + 10y) + (15x - 10y) = 3 + (-45) 6x = -42Solve for x: Step 4: Solve for x. 6x = -42 x = -42 / 6 x = -7Substitute and Find y: Step 5: Substitute x = -7 back into one of the original equations to find y. Use the second equation: 3x - 2y = -9. 3(-7) - 2y = -9 -21 - 2y = -9 -2y = 12 y = 12 / -2 y = -6

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Solve the system of equations. $\newline$ $-9x + 10y = 3$ $\newline$ $3x - 2y = -9$ $\newline$ (, )

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Precalculus

Solve a system of equations in three variables using elimination

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Q. Solve the system of equations.

\newline

-9x + 10y = 3

\newline

3x - 2y = -9

\newline

(____, ____)

Write Equations: Step $1$ : Write down the system of equations. $\newline$ $-9x + 10y = 3$ $\newline$ $3x - 2y = -9$
Eliminate y: Step $2$ : Use elimination method to eliminate y. Multiply the second equation by $5$ to balance the coefficients of $y$ . $\newline$ $5(3x - 2y) = 5(-9)$ $\newline$ $15x - 10y = -45$
Add Equations: Step $3$ : Add the modified second equation to the first equation to eliminate $y$ . $(-9x + 10y) + (15x - 10y) = 3 + (-45)$ $6x = -42$
Solve for x: Step $4$ : Solve for x. $\newline$ $6x = -42$ $\newline$ $x = \frac{-42}{6}$ $\newline$ $x = -7$
Substitute and Find y: Step $5$ : Substitute $x = -7$ back into one of the original equations to find $y$ . Use the second equation: $3x - 2y = -9$ . $3(-7) - 2y = -9$ $-21 - 2y = -9$ $-2y = 12$ $y = \frac{12}{-2}$ $y = -6$