Solve the system of equations. {:[5x-2y-4=0],[3x+16 y-54=0],[x=◻],[y=◻]:}

Isolate x in first equation: Isolate one of the variables in one of the equations. Let's isolate x in the first equation. 5x - 2y - 4 = 0 5x = 2y + 4 x = (2y + 4)/5Substitute x into second equation: Substitute the expression for x from Step 1 into the second equation. 3x + 16y - 54 = 0 3((2y + 4)/5) + 16y - 54 = 0 Multiply through by 5 to clear the fraction: 5[3((2y + 4)/5) + 16y - 54] = 5(0) 3(2y + 4) + 80y - 270 = 0Multiply through to clear fraction: Distribute and combine like terms. 6y + 12 + 80y - 270 = 0 86y - 258 = 0Combine like terms: Solve for y. 86y = 258 y = 258 / 86 y = 3Solve for y: Substitute y = 3 back into the expression for x from Step 1. x = (2y + 4)/5 x = (2(3) + 4)/5 x = (6 + 4)/5 x = 10/5 x = 2Substitute y back into x expression: Write the solution as a coordinate point. The solution is (2, 3).

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Solve the system of equations.

{:[5x-2y-4=0],[3x+16 y-54=0],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} 5 x-2 y-4=0 \\ 3 x+16 y-54=0 \\ x=\square \\ y=\square \end{array}$

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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} 5 x-2 y-4=0 \\ 3 x+16 y-54=0 \\ x=\square \\ y=\square \end{array}

Isolate $x$ in first equation: Isolate one of the variables in one of the equations. Let's isolate $x$ in the first equation. $\newline$ $5x - 2y - 4 = 0$ $\newline$ $5x = 2y + 4$ $\newline$ $x = \frac{2y + 4}{5}$
Substitute $x$ into second equation: Substitute the expression for $x$ from Step $1$ into the second equation. $3x + 16y - 54 = 0$ $3\left(\frac{2y + 4}{5}\right) + 16y - 54 = 0$ Multiply through by $5$ to clear the fraction: $5\left[3\left(\frac{2y + 4}{5}\right) + 16y - 54\right] = 5(0)$ $3(2y + 4) + 80y - 270 = 0$
Multiply through to clear fraction: Distribute and combine like terms. $\newline$ $6y + 12 + 80y - 270 = 0$ $\newline$ $86y - 258 = 0$
Combine like terms: Solve for $y$ . $86y = 258$ $y = \frac{258}{86}$ $y = 3$
Solve for y: Substitute $y = 3$ back into the expression for $x$ from Step $1$ . $\newline$ $x = \frac{2y + 4}{5}$ $\newline$ $x = \frac{2(3) + 4}{5}$ $\newline$ $x = \frac{6 + 4}{5}$ $\newline$ $x = \frac{10}{5}$ $\newline$ $x = 2$
Substitute $y$ back into $x$ expression: Write the solution as a coordinate point. $\newline$ The solution is $(2, 3)$ .