Solve {l}{text{i}x/10+2y/10+1=15}{x/8+y/6=15}

$$\left. \begin{array} { l } { \text { i } \frac { x } { 10 } + \frac { 2 y } { 10 } + 1 = 15 } \\ { \frac { x } { 8 } + \frac { y } { 6 } = 15 } \end{array} \right.$$

Solve for x, y

$$x=80$$

$$y=30$$

Steps Using Substitution

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

$$x+2y+10=150$$

Subtract 10 from both sides.

$$x+2y=150-10$$

Subtract 10 from 150 to get 140.

$$x+2y=140$$

Consider the second equation. Multiply both sides of the equation by 24, the least common multiple of 8,6.

$$3x+4y=360$$

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.

$$x+2y=140,3x+4y=360$$

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

$$x+2y=140$$

Subtract 2y from both sides of the equation.

$$x=-2y+140$$

Substitute -2y+140 for x in the other equation, 3x+4y=360.

$$3\left(-2y+140\right)+4y=360$$

Multiply 3 times -2y+140.

$$-6y+420+4y=360$$

Add -6y to 4y.

$$-2y+420=360$$

Subtract 420 from both sides of the equation.

$$-2y=-60$$

Divide both sides by -2.

$$y=30$$

Substitute 30 for y in x=-2y+140. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times 30+140$$

Multiply -2 times 30.

$$x=-60+140$$

Add 140 to -60.

$$x=80$$

The system is now solved.

$$x=80,y=30$$

Steps Using Matrices

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

$$x+2y+10=150$$

Subtract 10 from both sides.

$$x+2y=150-10$$

Subtract 10 from 150 to get 140.

$$x+2y=140$$

Consider the second equation. Multiply both sides of the equation by 24, the least common multiple of 8,6.

$$3x+4y=360$$

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

$$x+2y=140,3x+4y=360$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&2\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}140\\360\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\3&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&2\\3&4\end{matrix}\right))\left(\begin{matrix}1&2\\3&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\3&4\end{matrix}\right))\left(\begin{matrix}140\\360\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\3&4\end{matrix}\right))\left(\begin{matrix}140\\360\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\3&4\end{matrix}\right))\left(\begin{matrix}140\\360\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{4-2\times 3}&-\frac{2}{4-2\times 3}\\-\frac{3}{4-2\times 3}&\frac{1}{4-2\times 3}\end{matrix}\right)\left(\begin{matrix}140\\360\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&1\\\frac{3}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}140\\360\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\times 140+360\\\frac{3}{2}\times 140-\frac{1}{2}\times 360\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}80\\30\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=80,y=30$$

Steps Using Elimination

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

$$x+2y+10=150$$

Subtract 10 from both sides.

$$x+2y=150-10$$

Subtract 10 from 150 to get 140.

$$x+2y=140$$

Consider the second equation. Multiply both sides of the equation by 24, the least common multiple of 8,6.

$$3x+4y=360$$

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.

$$x+2y=140,3x+4y=360$$

To make x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 1.

$$3x+3\times 2y=3\times 140,3x+4y=360$$

Simplify.

$$3x+6y=420,3x+4y=360$$

Subtract 3x+4y=360 from 3x+6y=420 by subtracting like terms on each side of the equal sign.

$$3x-3x+6y-4y=420-360$$

Add 3x to -3x. Terms 3x and -3x cancel out, leaving an equation with only one variable that can be solved.

$$6y-4y=420-360$$

Add 6y to -4y.

$$2y=420-360$$

Add 420 to -360.

$$2y=60$$

Divide both sides by 2.

$$y=30$$

Substitute 30 for y in 3x+4y=360. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x+4\times 30=360$$

Multiply 4 times 30.

$$3x+120=360$$

Subtract 120 from both sides of the equation.

$$3x=240$$

Divide both sides by 3.

$$x=80$$

The system is now solved.

$$x=80,y=30$$