Solve {l}{y+3x=56}{x+4y=34}

$$\left. \begin{array} { l } { y + 3 x = 56 } \\ { x + 4 y = 34 } \end{array} \right.$$

Solve for y, x

$$x = \frac{190}{11} = 17\frac{3}{11} \approx 17.272727273$$

$$y = \frac{46}{11} = 4\frac{2}{11} \approx 4.181818182$$

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.

$$y+3x=56,4y+x=34$$

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

$$y+3x=56$$

Subtract 3x from both sides of the equation.

$$y=-3x+56$$

Substitute -3x+56 for y in the other equation, 4y+x=34.

$$4\left(-3x+56\right)+x=34$$

Multiply 4 times -3x+56.

$$-12x+224+x=34$$

Add -12x to x.

$$-11x+224=34$$

Subtract 224 from both sides of the equation.

$$-11x=-190$$

Divide both sides by -11.

$$x=\frac{190}{11}$$

Substitute \frac{190}{11} for x in y=-3x+56. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-3\times \frac{190}{11}+56$$

Multiply -3 times \frac{190}{11}.

$$y=-\frac{570}{11}+56$$

Add 56 to -\frac{570}{11}.

$$y=\frac{46}{11}$$

The system is now solved.

$$y=\frac{46}{11},x=\frac{190}{11}$$

Steps Using Matrices

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

$$y+3x=56,4y+x=34$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&3\\4&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}56\\34\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&3\\4&1\end{matrix}\right))\left(\begin{matrix}1&3\\4&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\4&1\end{matrix}\right))\left(\begin{matrix}56\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\4&1\end{matrix}\right))\left(\begin{matrix}56\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}1&3\\4&1\end{matrix}\right))\left(\begin{matrix}56\\34\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-3\times 4}&-\frac{3}{1-3\times 4}\\-\frac{4}{1-3\times 4}&\frac{1}{1-3\times 4}\end{matrix}\right)\left(\begin{matrix}56\\34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{11}&\frac{3}{11}\\\frac{4}{11}&-\frac{1}{11}\end{matrix}\right)\left(\begin{matrix}56\\34\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{11}\times 56+\frac{3}{11}\times 34\\\frac{4}{11}\times 56-\frac{1}{11}\times 34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{46}{11}\\\frac{190}{11}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{46}{11},x=\frac{190}{11}$$

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.

$$y+3x=56,4y+x=34$$

To make y and 4y equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.

$$4y+4\times 3x=4\times 56,4y+x=34$$

Simplify.

$$4y+12x=224,4y+x=34$$

Subtract 4y+x=34 from 4y+12x=224 by subtracting like terms on each side of the equal sign.

$$4y-4y+12x-x=224-34$$

Add 4y to -4y. Terms 4y and -4y cancel out, leaving an equation with only one variable that can be solved.

$$12x-x=224-34$$

Add 12x to -x.

$$11x=224-34$$

Add 224 to -34.

$$11x=190$$

Divide both sides by 11.

$$x=\frac{190}{11}$$

Substitute \frac{190}{11} for x in 4y+x=34. Because the resulting equation contains only one variable, you can solve for y directly.

$$4y+\frac{190}{11}=34$$

Subtract \frac{190}{11} from both sides of the equation.

$$4y=\frac{184}{11}$$

Divide both sides by 4.

$$y=\frac{46}{11}$$

The system is now solved.

$$y=\frac{46}{11},x=\frac{190}{11}$$