Solve {l}{2m-3n=130}{-m+5=4n}

$$\left\{ \begin{array} { l } { 2 m - 3 n = 130 } \\ { - m + 5 = 4 n } \end{array} \right.$$

Solve for m, n

$$m = \frac{535}{11} = 48\frac{7}{11} \approx 48.636363636$$

$$n = -\frac{120}{11} = -10\frac{10}{11} \approx -10.909090909$$

Steps Using Substitution

Consider the second equation. Subtract 4n from both sides.

$$-m+5-4n=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-m-4n=-5$$

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.

$$2m-3n=130,-m-4n=-5$$

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

$$2m-3n=130$$

Add 3n to both sides of the equation.

$$2m=3n+130$$

Divide both sides by 2.

$$m=\frac{1}{2}\left(3n+130\right)$$

Multiply \frac{1}{2} times 3n+130.

$$m=\frac{3}{2}n+65$$

Substitute \frac{3n}{2}+65 for m in the other equation, -m-4n=-5.

$$-\left(\frac{3}{2}n+65\right)-4n=-5$$

Multiply -1 times \frac{3n}{2}+65.

$$-\frac{3}{2}n-65-4n=-5$$

Add -\frac{3n}{2} to -4n.

$$-\frac{11}{2}n-65=-5$$

Add 65 to both sides of the equation.

$$-\frac{11}{2}n=60$$

Divide both sides of the equation by -\frac{11}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$n=-\frac{120}{11}$$

Substitute -\frac{120}{11} for n in m=\frac{3}{2}n+65. Because the resulting equation contains only one variable, you can solve for m directly.

$$m=\frac{3}{2}\left(-\frac{120}{11}\right)+65$$

Multiply \frac{3}{2} times -\frac{120}{11} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$m=-\frac{180}{11}+65$$

Add 65 to -\frac{180}{11}.

$$m=\frac{535}{11}$$

The system is now solved.

$$m=\frac{535}{11},n=-\frac{120}{11}$$

Steps Using Matrices

Consider the second equation. Subtract 4n from both sides.

$$-m+5-4n=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-m-4n=-5$$

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

$$2m-3n=130,-m-4n=-5$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}130\\-5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right))\left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right))\left(\begin{matrix}130\\-5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right))\left(\begin{matrix}130\\-5\end{matrix}\right)$$

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

$$\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\-1&-4\end{matrix}\right))\left(\begin{matrix}130\\-5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{2\left(-4\right)-\left(-3\left(-1\right)\right)}&-\frac{-3}{2\left(-4\right)-\left(-3\left(-1\right)\right)}\\-\frac{-1}{2\left(-4\right)-\left(-3\left(-1\right)\right)}&\frac{2}{2\left(-4\right)-\left(-3\left(-1\right)\right)}\end{matrix}\right)\left(\begin{matrix}130\\-5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{4}{11}&-\frac{3}{11}\\-\frac{1}{11}&-\frac{2}{11}\end{matrix}\right)\left(\begin{matrix}130\\-5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{4}{11}\times 130-\frac{3}{11}\left(-5\right)\\-\frac{1}{11}\times 130-\frac{2}{11}\left(-5\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{535}{11}\\-\frac{120}{11}\end{matrix}\right)$$

Extract the matrix elements m and n.

$$m=\frac{535}{11},n=-\frac{120}{11}$$

Steps Using Elimination

Consider the second equation. Subtract 4n from both sides.

$$-m+5-4n=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-m-4n=-5$$

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.

$$2m-3n=130,-m-4n=-5$$

To make 2m and -m equal, multiply all terms on each side of the first equation by -1 and all terms on each side of the second by 2.

$$-2m-\left(-3n\right)=-130,2\left(-1\right)m+2\left(-4\right)n=2\left(-5\right)$$

Simplify.

$$-2m+3n=-130,-2m-8n=-10$$

Subtract -2m-8n=-10 from -2m+3n=-130 by subtracting like terms on each side of the equal sign.

$$-2m+2m+3n+8n=-130+10$$

Add -2m to 2m. Terms -2m and 2m cancel out, leaving an equation with only one variable that can be solved.

$$3n+8n=-130+10$$

Add 3n to 8n.

$$11n=-130+10$$

Add -130 to 10.

$$11n=-120$$

Divide both sides by 11.

$$n=-\frac{120}{11}$$

Substitute -\frac{120}{11} for n in -m-4n=-5. Because the resulting equation contains only one variable, you can solve for m directly.

$$-m-4\left(-\frac{120}{11}\right)=-5$$

Multiply -4 times -\frac{120}{11}.

$$-m+\frac{480}{11}=-5$$

Subtract \frac{480}{11} from both sides of the equation.

$$-m=-\frac{535}{11}$$

Divide both sides by -1.

$$m=\frac{535}{11}$$

The system is now solved.

$$m=\frac{535}{11},n=-\frac{120}{11}$$