Solve the system of equations. {:[-2x+5y=13],[16 x+3y=25],[x=◻],[y=◻]:}

Multiply equation by 8: Multiply the first equation by 8 to make the coefficient of x in both equations the same. 8(-2x + 5y) = 8(13) -16x + 40y = 104Eliminate x: Add the new equation from Step 1 to the second equation to eliminate x. (-16x + 40y) + (16x + 3y) = 104 + 25 -16x + 40y + 16x + 3y = 129 43y = 129Solve for y: Solve for y by dividing both sides of the equation by 43. 43y / 43 = 129 / 43 y = 3Substitute y into first equation: Substitute y = 3 into the first original equation to solve for x. -2x + 5(3) = 13 -2x + 15 = 13 -2x = 13 - 15 -2x = -2Solve for x: Solve for x by dividing both sides of the equation by -2. -2x / -2 = -2 / -2 x = 1

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Solve the system of equations.

{:[-2x+5y=13],[16 x+3y=25],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} -2 x+5 y=13 \\ 16 x+3 y=25 \\ x=\square \\ y=\square \end{array}$

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Algebra 2

Solve a system of equations using elimination

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Q. Solve the system of equations.

\newline

\begin{array}{l} -2 x+5 y=13 \\ 16 x+3 y=25 \\ x=\square \\ y=\square \end{array}

Multiply equation by $8$ : Multiply the first equation by $8$ to make the coefficient of $x$ in both equations the same. $\newline$ $8(-2x + 5y) = 8(13)$ $\newline$ $-16x + 40y = 104$
Eliminate x: Add the new equation from Step $1$ to the second equation to eliminate x. $\newline$ $(-16x + 40y) + (16x + 3y) = 104 + 25$ $\newline$ $-16x + 40y + 16x + 3y = 129$ $\newline$ $43y = 129$
Solve for y: Solve for y by dividing both sides of the equation by $43$ . $\newline$ $\frac{43y}{43} = \frac{129}{43}$ $\newline$ $y = 3$
Substitute $y$ into first equation: Substitute $y = 3$ into the first original equation to solve for $x$ .
$-2x + 5(3) = 13$
$-2x + 15 = 13$
$-2x = 13 - 15$
$-2x = -2$
Solve for x: Solve for x by dividing both sides of the equation by \(-2＂). $\newline$ $-2x / -2 = -2 / -2$ $\newline$ $x = 1$