Solve {l}{5x=4y}{12x+2y=58}

$$\left\{ \begin{array} { l } { 5 x = 4 y } \\ { 12 x + 2 y = 58 } \end{array} \right.$$

Solve for x, y

$$x=4$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Subtract 4y from both sides.

$$5x-4y=0$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$5x-4y=0,12x+2y=58$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$5x-4y=0$$

Add 4y to both sides of the equation.

$$5x=4y$$

Divide both sides by 5.

$$x=\frac{1}{5}\times 4y$$

Multiply \frac{1}{5} times 4y.

$$x=\frac{4}{5}y$$

Substitute \frac{4y}{5} for x in the other equation, 12x+2y=58.

$$12\times \frac{4}{5}y+2y=58$$

Multiply 12 times \frac{4y}{5}.

$$\frac{48}{5}y+2y=58$$

Add \frac{48y}{5} to 2y.

$$\frac{58}{5}y=58$$

Divide both sides of the equation by \frac{58}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=5$$

Substitute 5 for y in x=\frac{4}{5}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{5}\times 5$$

Multiply \frac{4}{5} times 5.

$$x=4$$

The system is now solved.

$$x=4,y=5$$

Steps Using Matrices

Consider the first equation. Subtract 4y from both sides.

$$5x-4y=0$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$5x-4y=0,12x+2y=58$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-4\\12&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\58\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-4\\12&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-4\\12&2\end{matrix}\right))\left(\begin{matrix}5&-4\\12&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-4\\12&2\end{matrix}\right))\left(\begin{matrix}0\\58\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-4\\12&2\end{matrix}\right))\left(\begin{matrix}0\\58\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-4\\12&2\end{matrix}\right))\left(\begin{matrix}0\\58\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5\times 2-\left(-4\times 12\right)}&-\frac{-4}{5\times 2-\left(-4\times 12\right)}\\-\frac{12}{5\times 2-\left(-4\times 12\right)}&\frac{5}{5\times 2-\left(-4\times 12\right)}\end{matrix}\right)\left(\begin{matrix}0\\58\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{29}&\frac{2}{29}\\-\frac{6}{29}&\frac{5}{58}\end{matrix}\right)\left(\begin{matrix}0\\58\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{29}\times 58\\\frac{5}{58}\times 58\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4,y=5$$

Steps Using Elimination

Consider the first equation. Subtract 4y from both sides.

$$5x-4y=0$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$5x-4y=0,12x+2y=58$$

To make 5x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 5.

$$12\times 5x+12\left(-4\right)y=0,5\times 12x+5\times 2y=5\times 58$$

Simplify.

$$60x-48y=0,60x+10y=290$$

Subtract 60x+10y=290 from 60x-48y=0 by subtracting like terms on each side of the equal sign.

$$60x-60x-48y-10y=-290$$

Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.

$$-48y-10y=-290$$

Add -48y to -10y.

$$-58y=-290$$

Divide both sides by -58.

$$y=5$$

Substitute 5 for y in 12x+2y=58. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x+2\times 5=58$$

Multiply 2 times 5.

$$12x+10=58$$

Subtract 10 from both sides of the equation.

$$12x=48$$

Divide both sides by 12.

$$x=4$$

The system is now solved.

$$x=4,y=5$$