Solve {l}{x=1.5y}{60x=100+60y+20}

$$\left\{ \begin{array} { l } { x = 1.5 y } \\ { 60 x = 100 + 60 y + 20 } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=4$$

Steps Using Substitution

Consider the first equation. Subtract 1.5y from both sides.

$$x-1.5y=0$$

Consider the second equation. Add 100 and 20 to get 120.

$$60x=120+60y$$

Subtract 60y from both sides.

$$60x-60y=120$$

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.

$$x-1.5y=0,60x-60y=120$$

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

$$x-1.5y=0$$

Add \frac{3y}{2} to both sides of the equation.

$$x=1.5y$$

Substitute \frac{3y}{2} for x in the other equation, 60x-60y=120.

$$60\times 1.5y-60y=120$$

Multiply 60 times \frac{3y}{2}.

$$90y-60y=120$$

Add 90y to -60y.

$$30y=120$$

Divide both sides by 30.

$$y=4$$

Substitute 4 for y in x=1.5y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=1.5\times 4$$

Multiply 1.5 times 4.

$$x=6$$

The system is now solved.

$$x=6,y=4$$

Steps Using Matrices

Consider the first equation. Subtract 1.5y from both sides.

$$x-1.5y=0$$

Consider the second equation. Add 100 and 20 to get 120.

$$60x=120+60y$$

Subtract 60y from both sides.

$$60x-60y=120$$

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

$$x-1.5y=0,60x-60y=120$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\120\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right))\left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right))\left(\begin{matrix}0\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right))\left(\begin{matrix}0\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1.5\\60&-60\end{matrix}\right))\left(\begin{matrix}0\\120\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{60}{-60-\left(-1.5\times 60\right)}&-\frac{-1.5}{-60-\left(-1.5\times 60\right)}\\-\frac{60}{-60-\left(-1.5\times 60\right)}&\frac{1}{-60-\left(-1.5\times 60\right)}\end{matrix}\right)\left(\begin{matrix}0\\120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&\frac{1}{20}\\-2&\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}0\\120\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{20}\times 120\\\frac{1}{30}\times 120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=4$$

Steps Using Elimination

Consider the first equation. Subtract 1.5y from both sides.

$$x-1.5y=0$$

Consider the second equation. Add 100 and 20 to get 120.

$$60x=120+60y$$

Subtract 60y from both sides.

$$60x-60y=120$$

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.

$$x-1.5y=0,60x-60y=120$$

To make x and 60x equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 1.

$$60x+60\left(-1.5\right)y=0,60x-60y=120$$

Simplify.

$$60x-90y=0,60x-60y=120$$

Subtract 60x-60y=120 from 60x-90y=0 by subtracting like terms on each side of the equal sign.

$$60x-60x-90y+60y=-120$$

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

$$-90y+60y=-120$$

Add -90y to 60y.

$$-30y=-120$$

Divide both sides by -30.

$$y=4$$

Substitute 4 for y in 60x-60y=120. Because the resulting equation contains only one variable, you can solve for x directly.

$$60x-60\times 4=120$$

Multiply -60 times 4.

$$60x-240=120$$

Add 240 to both sides of the equation.

$$60x=360$$

Divide both sides by 60.

$$x=6$$

The system is now solved.

$$x=6,y=4$$