Solve {l}{30x=4/5y}{40x=y+25}

$$\left\{ \begin{array} { l } { 30 x = \frac { 4 } { 5 } y } \\ { 40 x = y + 25 } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=375$$

Steps Using Substitution

Consider the first equation. Subtract \frac{4}{5}y from both sides.

$$30x-\frac{4}{5}y=0$$

Consider the second equation. Subtract y from both sides.

$$40x-y=25$$

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.

$$30x-\frac{4}{5}y=0,40x-y=25$$

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

$$30x-\frac{4}{5}y=0$$

Add \frac{4y}{5} to both sides of the equation.

$$30x=\frac{4}{5}y$$

Divide both sides by 30.

$$x=\frac{1}{30}\times \frac{4}{5}y$$

Multiply \frac{1}{30} times \frac{4y}{5}.

$$x=\frac{2}{75}y$$

Substitute \frac{2y}{75} for x in the other equation, 40x-y=25.

$$40\times \frac{2}{75}y-y=25$$

Multiply 40 times \frac{2y}{75}.

$$\frac{16}{15}y-y=25$$

Add \frac{16y}{15} to -y.

$$\frac{1}{15}y=25$$

Multiply both sides by 15.

$$y=375$$

Substitute 375 for y in x=\frac{2}{75}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2}{75}\times 375$$

Multiply \frac{2}{75} times 375.

$$x=10$$

The system is now solved.

$$x=10,y=375$$

Steps Using Matrices

Consider the first equation. Subtract \frac{4}{5}y from both sides.

$$30x-\frac{4}{5}y=0$$

Consider the second equation. Subtract y from both sides.

$$40x-y=25$$

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

$$30x-\frac{4}{5}y=0,40x-y=25$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\25\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right))\left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right))\left(\begin{matrix}0\\25\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right))\left(\begin{matrix}0\\25\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&-\frac{4}{5}\\40&-1\end{matrix}\right))\left(\begin{matrix}0\\25\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{30\left(-1\right)-\left(-\frac{4}{5}\times 40\right)}&-\frac{-\frac{4}{5}}{30\left(-1\right)-\left(-\frac{4}{5}\times 40\right)}\\-\frac{40}{30\left(-1\right)-\left(-\frac{4}{5}\times 40\right)}&\frac{30}{30\left(-1\right)-\left(-\frac{4}{5}\times 40\right)}\end{matrix}\right)\left(\begin{matrix}0\\25\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&\frac{2}{5}\\-20&15\end{matrix}\right)\left(\begin{matrix}0\\25\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\times 25\\15\times 25\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\375\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=375$$

Steps Using Elimination

Consider the first equation. Subtract \frac{4}{5}y from both sides.

$$30x-\frac{4}{5}y=0$$

Consider the second equation. Subtract y from both sides.

$$40x-y=25$$

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.

$$30x-\frac{4}{5}y=0,40x-y=25$$

To make 30x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 30.

$$40\times 30x+40\left(-\frac{4}{5}\right)y=0,30\times 40x+30\left(-1\right)y=30\times 25$$

Simplify.

$$1200x-32y=0,1200x-30y=750$$

Subtract 1200x-30y=750 from 1200x-32y=0 by subtracting like terms on each side of the equal sign.

$$1200x-1200x-32y+30y=-750$$

Add 1200x to -1200x. Terms 1200x and -1200x cancel out, leaving an equation with only one variable that can be solved.

$$-32y+30y=-750$$

Add -32y to 30y.

$$-2y=-750$$

Divide both sides by -2.

$$y=375$$

Substitute 375 for y in 40x-y=25. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x-375=25$$

Add 375 to both sides of the equation.

$$40x=400$$

Divide both sides by 40.

$$x=10$$

The system is now solved.

$$x=10,y=375$$