Solve {l}{7y+38x=833}{2y+38x=713}

$$\left. \begin{array} { l } { 7 y + 38 x = 833 } \\ { 2 y + 38 x = 713 } \end{array} \right.$$

Solve for y, x

$$x = \frac{35}{2} = 17\frac{1}{2} = 17.5$$

$$y=24$$

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.

$$7y+38x=833,2y+38x=713$$

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

$$7y+38x=833$$

Subtract 38x from both sides of the equation.

$$7y=-38x+833$$

Divide both sides by 7.

$$y=\frac{1}{7}\left(-38x+833\right)$$

Multiply \frac{1}{7} times -38x+833.

$$y=-\frac{38}{7}x+119$$

Substitute -\frac{38x}{7}+119 for y in the other equation, 2y+38x=713.

$$2\left(-\frac{38}{7}x+119\right)+38x=713$$

Multiply 2 times -\frac{38x}{7}+119.

$$-\frac{76}{7}x+238+38x=713$$

Add -\frac{76x}{7} to 38x.

$$\frac{190}{7}x+238=713$$

Subtract 238 from both sides of the equation.

$$\frac{190}{7}x=475$$

Divide both sides of the equation by \frac{190}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{35}{2}$$

Substitute \frac{35}{2} for x in y=-\frac{38}{7}x+119. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{38}{7}\times \frac{35}{2}+119$$

Multiply -\frac{38}{7} times \frac{35}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-95+119$$

Add 119 to -95.

$$y=24$$

The system is now solved.

$$y=24,x=\frac{35}{2}$$

Steps Using Matrices

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

$$7y+38x=833,2y+38x=713$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&38\\2&38\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}833\\713\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&38\\2&38\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&38\\2&38\end{matrix}\right))\left(\begin{matrix}7&38\\2&38\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&38\\2&38\end{matrix}\right))\left(\begin{matrix}833\\713\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&38\\2&38\end{matrix}\right))\left(\begin{matrix}833\\713\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&38\\2&38\end{matrix}\right))\left(\begin{matrix}833\\713\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{38}{7\times 38-38\times 2}&-\frac{38}{7\times 38-38\times 2}\\-\frac{2}{7\times 38-38\times 2}&\frac{7}{7\times 38-38\times 2}\end{matrix}\right)\left(\begin{matrix}833\\713\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&-\frac{1}{5}\\-\frac{1}{95}&\frac{7}{190}\end{matrix}\right)\left(\begin{matrix}833\\713\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\times 833-\frac{1}{5}\times 713\\-\frac{1}{95}\times 833+\frac{7}{190}\times 713\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}24\\\frac{35}{2}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=24,x=\frac{35}{2}$$

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.

$$7y+38x=833,2y+38x=713$$

Subtract 2y+38x=713 from 7y+38x=833 by subtracting like terms on each side of the equal sign.

$$7y-2y+38x-38x=833-713$$

Add 38x to -38x. Terms 38x and -38x cancel out, leaving an equation with only one variable that can be solved.

$$7y-2y=833-713$$

Add 7y to -2y.

$$5y=833-713$$

Add 833 to -713.

$$5y=120$$

Divide both sides by 5.

$$y=24$$

Substitute 24 for y in 2y+38x=713. Because the resulting equation contains only one variable, you can solve for x directly.

$$2\times 24+38x=713$$

Multiply 2 times 24.

$$48+38x=713$$

Subtract 48 from both sides of the equation.

$$38x=665$$

Divide both sides by 38.

$$x=\frac{35}{2}$$

The system is now solved.

$$y=24,x=\frac{35}{2}$$