Solve {l}{12x-5y=40}{12x-11y=88}

$$\left. \begin{array} { l } { 12 x - 5 y = 40 } \\ { 12 x - 11 y = 88 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=-8$$

Steps Using Substitution

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.

$$12x-5y=40,12x-11y=88$$

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

$$12x-5y=40$$

Add 5y to both sides of the equation.

$$12x=5y+40$$

Divide both sides by 12.

$$x=\frac{1}{12}\left(5y+40\right)$$

Multiply \frac{1}{12} times 40+5y.

$$x=\frac{5}{12}y+\frac{10}{3}$$

Substitute \frac{10}{3}+\frac{5y}{12} for x in the other equation, 12x-11y=88.

$$12\left(\frac{5}{12}y+\frac{10}{3}\right)-11y=88$$

Multiply 12 times \frac{10}{3}+\frac{5y}{12}.

$$5y+40-11y=88$$

Add 5y to -11y.

$$-6y+40=88$$

Subtract 40 from both sides of the equation.

$$-6y=48$$

Divide both sides by -6.

$$y=-8$$

Substitute -8 for y in x=\frac{5}{12}y+\frac{10}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{12}\left(-8\right)+\frac{10}{3}$$

Multiply \frac{5}{12} times -8.

$$x=\frac{-10+10}{3}$$

Add \frac{10}{3} to -\frac{10}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=0$$

The system is now solved.

$$x=0,y=-8$$

Steps Using Matrices

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

$$12x-5y=40,12x-11y=88$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&-5\\12&-11\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}40\\88\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&-5\\12&-11\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&-5\\12&-11\end{matrix}\right))\left(\begin{matrix}12&-5\\12&-11\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&-5\\12&-11\end{matrix}\right))\left(\begin{matrix}40\\88\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&-5\\12&-11\end{matrix}\right))\left(\begin{matrix}40\\88\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&-5\\12&-11\end{matrix}\right))\left(\begin{matrix}40\\88\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{11}{12\left(-11\right)-\left(-5\times 12\right)}&-\frac{-5}{12\left(-11\right)-\left(-5\times 12\right)}\\-\frac{12}{12\left(-11\right)-\left(-5\times 12\right)}&\frac{12}{12\left(-11\right)-\left(-5\times 12\right)}\end{matrix}\right)\left(\begin{matrix}40\\88\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{72}&-\frac{5}{72}\\\frac{1}{6}&-\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}40\\88\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{11}{72}\times 40-\frac{5}{72}\times 88\\\frac{1}{6}\times 40-\frac{1}{6}\times 88\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-8\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=-8$$

Steps Using Elimination

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.

$$12x-5y=40,12x-11y=88$$

Subtract 12x-11y=88 from 12x-5y=40 by subtracting like terms on each side of the equal sign.

$$12x-12x-5y+11y=40-88$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$-5y+11y=40-88$$

Add -5y to 11y.

$$6y=40-88$$

Add 40 to -88.

$$6y=-48$$

Divide both sides by 6.

$$y=-8$$

Substitute -8 for y in 12x-11y=88. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x-11\left(-8\right)=88$$

Multiply -11 times -8.

$$12x+88=88$$

Subtract 88 from both sides of the equation.

$$12x=0$$

Divide both sides by 12.

$$x=0$$

The system is now solved.

$$x=0,y=-8$$