Solve the system of equations. {:[10 y+7x=29],[-5y-9x=2],[x=◻],[y=◻]:}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can choose to eliminate 'y' by multiplying the first equation by 5 and the second equation by 10 to make the coefficients of 'y' in both equations equal and opposite.Multiply equations by coefficients: Multiply the first equation by 5 and the second equation by 10. First equation: 5(10y + 7x) = 5(29) gives 50y + 35x = 145. Second equation: 10(-5y - 9x) = 10(2) gives -50y - 90x = 20.Add equations to eliminate variable: Add the equations to eliminate 'y'. (50y + 35x) + (-50y - 90x) = 145 + 20 50y - 50y + 35x - 90x = 165 0y - 55x = 165 -55x = 165Solve for x: Solve for 'x'. Dividing both sides of the equation by -55 gives us x = -3.Substitute x into first equation: Substitute x = -3 into the first equation to solve for 'y'. Substitute x = -3 in 10y + 7x = 29. We get 10y + 7(-3) = 29. 10y - 21 = 29. Add 21 to both sides, we get 10y = 50. Divide by 10, we get y = 5.Write solution as coordinate point: Write the solution as a coordinate point. The solution is (-3, 5).

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Solve the system of equations.

{:[10 y+7x=29],[-5y-9x=2],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} 10 y+7 x=29 \\ -5 y-9 x=2 \\ x=\square \\ y=\square \end{array}$

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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} 10 y+7 x=29 \\ -5 y-9 x=2 \\ x=\square \\ y=\square \end{array}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can choose to eliminate ' $y$ ' by multiplying the first equation by $5$ and the second equation by $10$ to make the coefficients of ' $y$ ' in both equations equal and opposite.
Multiply equations by coefficients: Multiply the first equation by $5$ and the second equation by $10$ .
First equation: $5(10y + 7x) = 5(29)$ gives $50y + 35x = 145$ .
Second equation: $10(-5y - 9x) = 10(2)$ gives $-50y - 90x = 20$ .
Add equations to eliminate variable: Add the equations to eliminate $y$ .
$(50y + 35x) + (-50y - 90x) = 145 + 20$
$50y - 50y + 35x - 90x = 165$
$0y - 55x = 165$
$-55x = 165$
Solve for x: Solve for 'x'. Dividing both sides of the equation by $-55$ gives us $x = -3$ .
Substitute $x$ into first equation: Substitute $x = -3$ into the first equation to solve for 'y'. $\newline$ Substitute $x = -3$ in $10y + 7x = 29$ . We get $10y + 7(-3) = 29$ . $\newline$ $10y - 21 = 29$ . Add $21$ to both sides, we get $10y = 50$ . $\newline$ Divide by $10$ , we get $y = 5$ .
Write solution as coordinate point: Write the solution as a coordinate point. The solution is $(-3, 5)$ .