Solve {l}{5x=20y-25}{-5x=10y-30}

$$\left. \begin{array} { l } { 5 x = 20 y - 25 } \\ { - 5 x = 10 y - 30 } \end{array} \right.$$

Solve for x, y

$$x = \frac{7}{3} = 2\frac{1}{3} \approx 2.333333333$$

$$y = \frac{11}{6} = 1\frac{5}{6} \approx 1.833333333$$

Steps Using Substitution

Consider the first equation. Subtract 20y from both sides.

$$5x-20y=-25$$

Consider the second equation. Subtract 10y from both sides.

$$-5x-10y=-30$$

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.

$$5x-20y=-25,-5x-10y=-30$$

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

$$5x-20y=-25$$

Add 20y to both sides of the equation.

$$5x=20y-25$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(20y-25\right)$$

Multiply \frac{1}{5} times 20y-25.

$$x=4y-5$$

Substitute 4y-5 for x in the other equation, -5x-10y=-30.

$$-5\left(4y-5\right)-10y=-30$$

Multiply -5 times 4y-5.

$$-20y+25-10y=-30$$

Add -20y to -10y.

$$-30y+25=-30$$

Subtract 25 from both sides of the equation.

$$-30y=-55$$

Divide both sides by -30.

$$y=\frac{11}{6}$$

Substitute \frac{11}{6} for y in x=4y-5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=4\times \frac{11}{6}-5$$

Multiply 4 times \frac{11}{6}.

$$x=\frac{22}{3}-5$$

Add -5 to \frac{22}{3}.

$$x=\frac{7}{3}$$

The system is now solved.

$$x=\frac{7}{3},y=\frac{11}{6}$$

Steps Using Matrices

Consider the first equation. Subtract 20y from both sides.

$$5x-20y=-25$$

Consider the second equation. Subtract 10y from both sides.

$$-5x-10y=-30$$

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

$$5x-20y=-25,-5x-10y=-30$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-25\\-30\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right))\left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right))\left(\begin{matrix}-25\\-30\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right))\left(\begin{matrix}-25\\-30\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-20\\-5&-10\end{matrix}\right))\left(\begin{matrix}-25\\-30\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{10}{5\left(-10\right)-\left(-20\left(-5\right)\right)}&-\frac{-20}{5\left(-10\right)-\left(-20\left(-5\right)\right)}\\-\frac{-5}{5\left(-10\right)-\left(-20\left(-5\right)\right)}&\frac{5}{5\left(-10\right)-\left(-20\left(-5\right)\right)}\end{matrix}\right)\left(\begin{matrix}-25\\-30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{15}&-\frac{2}{15}\\-\frac{1}{30}&-\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}-25\\-30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{15}\left(-25\right)-\frac{2}{15}\left(-30\right)\\-\frac{1}{30}\left(-25\right)-\frac{1}{30}\left(-30\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{3}\\\frac{11}{6}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{7}{3},y=\frac{11}{6}$$

Steps Using Elimination

Consider the first equation. Subtract 20y from both sides.

$$5x-20y=-25$$

Consider the second equation. Subtract 10y from both sides.

$$-5x-10y=-30$$

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.

$$5x-20y=-25,-5x-10y=-30$$

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

$$-5\times 5x-5\left(-20\right)y=-5\left(-25\right),5\left(-5\right)x+5\left(-10\right)y=5\left(-30\right)$$

Simplify.

$$-25x+100y=125,-25x-50y=-150$$

Subtract -25x-50y=-150 from -25x+100y=125 by subtracting like terms on each side of the equal sign.

$$-25x+25x+100y+50y=125+150$$

Add -25x to 25x. Terms -25x and 25x cancel out, leaving an equation with only one variable that can be solved.

$$100y+50y=125+150$$

Add 100y to 50y.

$$150y=125+150$$

Add 125 to 150.

$$150y=275$$

Divide both sides by 150.

$$y=\frac{11}{6}$$

Substitute \frac{11}{6} for y in -5x-10y=-30. Because the resulting equation contains only one variable, you can solve for x directly.

$$-5x-10\times \frac{11}{6}=-30$$

Multiply -10 times \frac{11}{6}.

$$-5x-\frac{55}{3}=-30$$

Add \frac{55}{3} to both sides of the equation.

$$-5x=-\frac{35}{3}$$

Divide both sides by -5.

$$x=\frac{7}{3}$$

The system is now solved.

$$x=\frac{7}{3},y=\frac{11}{6}$$