Solve {l}{2y+7x/10=18}{7x+18y=30}

$$\left. \begin{array} { l } { 2 y + \frac { 7 x } { 10 } = 18 } \\ { 7 x + 18 y = 30 } \end{array} \right.$$

Solve for y, x

$$x = -\frac{1320}{7} = -188\frac{4}{7} \approx -188.571428571$$

$$y=75$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 10.

$$20y+7x=180$$

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.

$$20y+7x=180,18y+7x=30$$

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

$$20y+7x=180$$

Subtract 7x from both sides of the equation.

$$20y=-7x+180$$

Divide both sides by 20.

$$y=\frac{1}{20}\left(-7x+180\right)$$

Multiply \frac{1}{20} times -7x+180.

$$y=-\frac{7}{20}x+9$$

Substitute -\frac{7x}{20}+9 for y in the other equation, 18y+7x=30.

$$18\left(-\frac{7}{20}x+9\right)+7x=30$$

Multiply 18 times -\frac{7x}{20}+9.

$$-\frac{63}{10}x+162+7x=30$$

Add -\frac{63x}{10} to 7x.

$$\frac{7}{10}x+162=30$$

Subtract 162 from both sides of the equation.

$$\frac{7}{10}x=-132$$

Divide both sides of the equation by \frac{7}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=-\frac{1320}{7}$$

Substitute -\frac{1320}{7} for x in y=-\frac{7}{20}x+9. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{7}{20}\left(-\frac{1320}{7}\right)+9$$

Multiply -\frac{7}{20} times -\frac{1320}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=66+9$$

Add 9 to 66.

$$y=75$$

The system is now solved.

$$y=75,x=-\frac{1320}{7}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 10.

$$20y+7x=180$$

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

$$20y+7x=180,18y+7x=30$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&7\\18&7\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}180\\30\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&7\\18&7\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&7\\18&7\end{matrix}\right))\left(\begin{matrix}20&7\\18&7\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}20&7\\18&7\end{matrix}\right))\left(\begin{matrix}180\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}20&7\\18&7\end{matrix}\right))\left(\begin{matrix}180\\30\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}20&7\\18&7\end{matrix}\right))\left(\begin{matrix}180\\30\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{7}{20\times 7-7\times 18}&-\frac{7}{20\times 7-7\times 18}\\-\frac{18}{20\times 7-7\times 18}&\frac{20}{20\times 7-7\times 18}\end{matrix}\right)\left(\begin{matrix}180\\30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&-\frac{1}{2}\\-\frac{9}{7}&\frac{10}{7}\end{matrix}\right)\left(\begin{matrix}180\\30\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 180-\frac{1}{2}\times 30\\-\frac{9}{7}\times 180+\frac{10}{7}\times 30\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}75\\-\frac{1320}{7}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=75,x=-\frac{1320}{7}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 10.

$$20y+7x=180$$

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.

$$20y+7x=180,18y+7x=30$$

Subtract 18y+7x=30 from 20y+7x=180 by subtracting like terms on each side of the equal sign.

$$20y-18y+7x-7x=180-30$$

Add 7x to -7x. Terms 7x and -7x cancel out, leaving an equation with only one variable that can be solved.

$$20y-18y=180-30$$

Add 20y to -18y.

$$2y=180-30$$

Add 180 to -30.

$$2y=150$$

Divide both sides by 2.

$$y=75$$

Substitute 75 for y in 18y+7x=30. Because the resulting equation contains only one variable, you can solve for x directly.

$$18\times 75+7x=30$$

Multiply 18 times 75.

$$1350+7x=30$$

Subtract 1350 from both sides of the equation.

$$7x=-1320$$

Divide both sides by 7.

$$x=-\frac{1320}{7}$$

The system is now solved.

$$y=75,x=-\frac{1320}{7}$$