Solve {l}{11x+5y-58=0}{14x-8y-2=0}

$$\left. \begin{array} { l } { 11 x + 5 y - 58 = 0 } \\ { 14 x - 8 y - 2 = 0 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=5$$

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.

$$11x+5y-58=0,14x-8y-2=0$$

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

$$11x+5y-58=0$$

Add 58 to both sides of the equation.

$$11x+5y=58$$

Subtract 5y from both sides of the equation.

$$11x=-5y+58$$

Divide both sides by 11.

$$x=\frac{1}{11}\left(-5y+58\right)$$

Multiply \frac{1}{11} times -5y+58.

$$x=-\frac{5}{11}y+\frac{58}{11}$$

Substitute \frac{-5y+58}{11} for x in the other equation, 14x-8y-2=0.

$$14\left(-\frac{5}{11}y+\frac{58}{11}\right)-8y-2=0$$

Multiply 14 times \frac{-5y+58}{11}.

$$-\frac{70}{11}y+\frac{812}{11}-8y-2=0$$

Add -\frac{70y}{11} to -8y.

$$-\frac{158}{11}y+\frac{812}{11}-2=0$$

Add \frac{812}{11} to -2.

$$-\frac{158}{11}y+\frac{790}{11}=0$$

Subtract \frac{790}{11} from both sides of the equation.

$$-\frac{158}{11}y=-\frac{790}{11}$$

Divide both sides of the equation by -\frac{158}{11}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=5$$

Substitute 5 for y in x=-\frac{5}{11}y+\frac{58}{11}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{11}\times 5+\frac{58}{11}$$

Multiply -\frac{5}{11} times 5.

$$x=\frac{-25+58}{11}$$

Add \frac{58}{11} to -\frac{25}{11} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=3$$

The system is now solved.

$$x=3,y=5$$

Steps Using Matrices

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

$$11x+5y-58=0,14x-8y-2=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}11&5\\14&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}58\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}11&5\\14&-8\end{matrix}\right).

$$inverse(\left(\begin{matrix}11&5\\14&-8\end{matrix}\right))\left(\begin{matrix}11&5\\14&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}11&5\\14&-8\end{matrix}\right))\left(\begin{matrix}58\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}11&5\\14&-8\end{matrix}\right))\left(\begin{matrix}58\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}11&5\\14&-8\end{matrix}\right))\left(\begin{matrix}58\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{11\left(-8\right)-5\times 14}&-\frac{5}{11\left(-8\right)-5\times 14}\\-\frac{14}{11\left(-8\right)-5\times 14}&\frac{11}{11\left(-8\right)-5\times 14}\end{matrix}\right)\left(\begin{matrix}58\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{79}&\frac{5}{158}\\\frac{7}{79}&-\frac{11}{158}\end{matrix}\right)\left(\begin{matrix}58\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{79}\times 58+\frac{5}{158}\times 2\\\frac{7}{79}\times 58-\frac{11}{158}\times 2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=5$$

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.

$$11x+5y-58=0,14x-8y-2=0$$

To make 11x and 14x equal, multiply all terms on each side of the first equation by 14 and all terms on each side of the second by 11.

$$14\times 11x+14\times 5y+14\left(-58\right)=0,11\times 14x+11\left(-8\right)y+11\left(-2\right)=0$$

Simplify.

$$154x+70y-812=0,154x-88y-22=0$$

Subtract 154x-88y-22=0 from 154x+70y-812=0 by subtracting like terms on each side of the equal sign.

$$154x-154x+70y+88y-812+22=0$$

Add 154x to -154x. Terms 154x and -154x cancel out, leaving an equation with only one variable that can be solved.

$$70y+88y-812+22=0$$

Add 70y to 88y.

$$158y-812+22=0$$

Add -812 to 22.

$$158y-790=0$$

Add 790 to both sides of the equation.

$$158y=790$$

Divide both sides by 158.

$$y=5$$

Substitute 5 for y in 14x-8y-2=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$14x-8\times 5-2=0$$

Multiply -8 times 5.

$$14x-40-2=0$$

Add -40 to -2.

$$14x-42=0$$

Add 42 to both sides of the equation.

$$14x=42$$

Divide both sides by 14.

$$x=3$$

The system is now solved.

$$x=3,y=5$$