Solve using elimination. –10x + 5y = –15 –3x + 5y = 20 (_____, _____)

Write Equations: Write down the system of equations to be solved using elimination. –10x + 5y = –15 –3x + 5y = 20Eliminate Variable: To use elimination, we want to eliminate one of the variables by adding or subtracting the equations. Since the coefficients of y are the same in both equations, we can subtract the second equation from the first to eliminate y. Subtract the second equation from the first: (–10x + 5y) - (–3x + 5y) = –15 - 20Solve for x: Perform the subtraction to eliminate y and solve for x. –10x + 5y - (–3x) - 5y = –15 - 20 –10x + 3x = –35 –7x = –35Substitute x: Divide both sides by –7 to solve for x. x = –35 / –7 x = 5Solve for y: Now that we have the value of x, we can substitute it back into one of the original equations to solve for y. Let's use the first equation: –10x + 5y = –15 Substitute x = 5 into the equation: –10(5) + 5y = –15Final Solution: Perform the multiplication and solve for y. –50 + 5y = –15 5y = –15 + 50 5y = 35Divide both sides by 5 to solve for y. y = 35 / 5 y = 7

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Solve using elimination. $\newline$ $-10x + 5y = -15$ $\newline$ $-3x + 5y = 20$ $\newline$ (_, _)

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Grade 8

Solve a system of equations using elimination

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Q. Solve using elimination.

\newline

-10x + 5y = -15

\newline

-3x + 5y = 20

\newline

(_____, _____)

Write Equations: Write down the system of equations to be solved using elimination. $\newline$ $-10x + 5y = -15$ $\newline$ $-3x + 5y = 20$
Eliminate Variable: To use elimination, we want to eliminate one of the variables by adding or subtracting the equations. Since the coefficients of $y$ are the same in both equations, we can subtract the second equation from the first to eliminate $y$ . $\newline$ Subtract the second equation from the first: $\newline$ $(–10x + 5y) - (–3x + 5y) = –15 - 20$
Solve for x: Perform the subtraction to eliminate $y$ and solve for $x$ .
$-10x + 5y - (-3x) - 5y = -15 - 20$
$-10x + 3x = -35$
$-7x = -35$
Substitute x: Divide both sides by $–7$ to solve for x. $\newline$ $x = \frac{–35}{–7}$ $\newline$ $x = 5$
Solve for y: Now that we have the value of $x$ , we can substitute it back into one of the original equations to solve for $y$ . Let's use the first equation: $\newline$ $–10x + 5y = –15$ $\newline$ Substitute $x = 5$ into the equation: $\newline$ $–10(5) + 5y = –15$
Final Solution: Perform the multiplication and solve for $y$ . $-50 + 5y = -15$ $5y = -15 + 50$ $5y = 35$
Final Solution: Perform the multiplication and solve for $y$ . $-50 + 5y = -15$ $5y = -15 + 50$ $5y = 35$ Divide both sides by $5$ to solve for $y$ . $y = \frac{35}{5}$ $y = 7$