Solve {l}{5y=2x}{2(2x+x+2y)=76}

$$\left\{ \begin{array} { l } { 5 y = 2 x } \\ { 2 ( 2 x + x + 2 y ) = 76 } \end{array} \right.$$

Solve for y, x

$$x=10$$

$$y=4$$

Steps Using Substitution

Consider the first equation. Subtract 2x from both sides.

$$5y-2x=0$$

Consider the second equation. Divide both sides by 2.

$$2x+x+2y=\frac{76}{2}$$

Divide 76 by 2 to get 38.

$$2x+x+2y=38$$

Combine 2x and x to get 3x.

$$3x+2y=38$$

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.

$$5y-2x=0,2y+3x=38$$

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

$$5y-2x=0$$

Add 2x to both sides of the equation.

$$5y=2x$$

Divide both sides by 5.

$$y=\frac{1}{5}\times 2x$$

Multiply \frac{1}{5} times 2x.

$$y=\frac{2}{5}x$$

Substitute \frac{2x}{5} for y in the other equation, 2y+3x=38.

$$2\times \frac{2}{5}x+3x=38$$

Multiply 2 times \frac{2x}{5}.

$$\frac{4}{5}x+3x=38$$

Add \frac{4x}{5} to 3x.

$$\frac{19}{5}x=38$$

Divide both sides of the equation by \frac{19}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=10$$

Substitute 10 for x in y=\frac{2}{5}x. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{2}{5}\times 10$$

Multiply \frac{2}{5} times 10.

$$y=4$$

The system is now solved.

$$y=4,x=10$$

Steps Using Matrices

Consider the first equation. Subtract 2x from both sides.

$$5y-2x=0$$

Consider the second equation. Divide both sides by 2.

$$2x+x+2y=\frac{76}{2}$$

Divide 76 by 2 to get 38.

$$2x+x+2y=38$$

Combine 2x and x to get 3x.

$$3x+2y=38$$

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

$$5y-2x=0,2y+3x=38$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-2\\2&3\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\38\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-2\\2&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-2\\2&3\end{matrix}\right))\left(\begin{matrix}5&-2\\2&3\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\2&3\end{matrix}\right))\left(\begin{matrix}0\\38\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\2&3\end{matrix}\right))\left(\begin{matrix}0\\38\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\2&3\end{matrix}\right))\left(\begin{matrix}0\\38\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5\times 3-\left(-2\times 2\right)}&-\frac{-2}{5\times 3-\left(-2\times 2\right)}\\-\frac{2}{5\times 3-\left(-2\times 2\right)}&\frac{5}{5\times 3-\left(-2\times 2\right)}\end{matrix}\right)\left(\begin{matrix}0\\38\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{3}{19}&\frac{2}{19}\\-\frac{2}{19}&\frac{5}{19}\end{matrix}\right)\left(\begin{matrix}0\\38\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{19}\times 38\\\frac{5}{19}\times 38\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}4\\10\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=4,x=10$$

Steps Using Elimination

Consider the first equation. Subtract 2x from both sides.

$$5y-2x=0$$

Consider the second equation. Divide both sides by 2.

$$2x+x+2y=\frac{76}{2}$$

Divide 76 by 2 to get 38.

$$2x+x+2y=38$$

Combine 2x and x to get 3x.

$$3x+2y=38$$

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.

$$5y-2x=0,2y+3x=38$$

To make 5y and 2y equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 5.

$$2\times 5y+2\left(-2\right)x=0,5\times 2y+5\times 3x=5\times 38$$

Simplify.

$$10y-4x=0,10y+15x=190$$

Subtract 10y+15x=190 from 10y-4x=0 by subtracting like terms on each side of the equal sign.

$$10y-10y-4x-15x=-190$$

Add 10y to -10y. Terms 10y and -10y cancel out, leaving an equation with only one variable that can be solved.

$$-4x-15x=-190$$

Add -4x to -15x.

$$-19x=-190$$

Divide both sides by -19.

$$x=10$$

Substitute 10 for x in 2y+3x=38. Because the resulting equation contains only one variable, you can solve for y directly.

$$2y+3\times 10=38$$

Multiply 3 times 10.

$$2y+30=38$$

Subtract 30 from both sides of the equation.

$$2y=8$$

Divide both sides by 2.

$$y=4$$

The system is now solved.

$$y=4,x=10$$