Solve {l}{2y+7x=30}{10y+x=40}

$$\left\{ \begin{array} { l } { 2 y + 7 x = 30 } \\ { 10 y + x = 40 } \end{array} \right.$$

Solve for y, x

$$x = \frac{55}{17} = 3\frac{4}{17} \approx 3.235294118$$

$$y = \frac{125}{34} = 3\frac{23}{34} \approx 3.676470588$$

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.

$$2y+7x=30,10y+x=40$$

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

$$2y+7x=30$$

Subtract 7x from both sides of the equation.

$$2y=-7x+30$$

Divide both sides by 2.

$$y=\frac{1}{2}\left(-7x+30\right)$$

Multiply \frac{1}{2} times -7x+30.

$$y=-\frac{7}{2}x+15$$

Substitute -\frac{7x}{2}+15 for y in the other equation, 10y+x=40.

$$10\left(-\frac{7}{2}x+15\right)+x=40$$

Multiply 10 times -\frac{7x}{2}+15.

$$-35x+150+x=40$$

Add -35x to x.

$$-34x+150=40$$

Subtract 150 from both sides of the equation.

$$-34x=-110$$

Divide both sides by -34.

$$x=\frac{55}{17}$$

Substitute \frac{55}{17} for x in y=-\frac{7}{2}x+15. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{7}{2}\times \frac{55}{17}+15$$

Multiply -\frac{7}{2} times \frac{55}{17} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=-\frac{385}{34}+15$$

Add 15 to -\frac{385}{34}.

$$y=\frac{125}{34}$$

The system is now solved.

$$y=\frac{125}{34},x=\frac{55}{17}$$

Steps Using Matrices

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

$$2y+7x=30,10y+x=40$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&7\\10&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}30\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&7\\10&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&7\\10&1\end{matrix}\right))\left(\begin{matrix}2&7\\10&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&7\\10&1\end{matrix}\right))\left(\begin{matrix}30\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&7\\10&1\end{matrix}\right))\left(\begin{matrix}30\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&7\\10&1\end{matrix}\right))\left(\begin{matrix}30\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2-7\times 10}&-\frac{7}{2-7\times 10}\\-\frac{10}{2-7\times 10}&\frac{2}{2-7\times 10}\end{matrix}\right)\left(\begin{matrix}30\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{68}&\frac{7}{68}\\\frac{5}{34}&-\frac{1}{34}\end{matrix}\right)\left(\begin{matrix}30\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{68}\times 30+\frac{7}{68}\times 40\\\frac{5}{34}\times 30-\frac{1}{34}\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{125}{34}\\\frac{55}{17}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{125}{34},x=\frac{55}{17}$$

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.

$$2y+7x=30,10y+x=40$$

To make 2y and 10y equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 2.

$$10\times 2y+10\times 7x=10\times 30,2\times 10y+2x=2\times 40$$

Simplify.

$$20y+70x=300,20y+2x=80$$

Subtract 20y+2x=80 from 20y+70x=300 by subtracting like terms on each side of the equal sign.

$$20y-20y+70x-2x=300-80$$

Add 20y to -20y. Terms 20y and -20y cancel out, leaving an equation with only one variable that can be solved.

$$70x-2x=300-80$$

Add 70x to -2x.

$$68x=300-80$$

Add 300 to -80.

$$68x=220$$

Divide both sides by 68.

$$x=\frac{55}{17}$$

Substitute \frac{55}{17} for x in 10y+x=40. Because the resulting equation contains only one variable, you can solve for y directly.

$$10y+\frac{55}{17}=40$$

Subtract \frac{55}{17} from both sides of the equation.

$$10y=\frac{625}{17}$$

Divide both sides by 10.

$$y=\frac{125}{34}$$

The system is now solved.

$$y=\frac{125}{34},x=\frac{55}{17}$$