Solve {l}{x/2+y/3=10}{5x/7-2y/9=8}

$$\left. \begin{array} { l } { \frac { x } { 2 } + \frac { y } { 3 } = 10 } \\ { \frac { 5 x } { 7 } - \frac { 2 y } { 9 } = 8 } \end{array} \right.$$

Solve for x, y

$$x=14$$

$$y=9$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x+2y=60$$

Consider the second equation. Multiply both sides of the equation by 63, the least common multiple of 7,9.

$$9\times 5x-7\times 2y=504$$

Multiply 9 and 5 to get 45.

$$45x-7\times 2y=504$$

Multiply -7 and 2 to get -14.

$$45x-14y=504$$

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.

$$3x+2y=60,45x-14y=504$$

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

$$3x+2y=60$$

Subtract 2y from both sides of the equation.

$$3x=-2y+60$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-2y+60\right)$$

Multiply \frac{1}{3} times -2y+60.

$$x=-\frac{2}{3}y+20$$

Substitute -\frac{2y}{3}+20 for x in the other equation, 45x-14y=504.

$$45\left(-\frac{2}{3}y+20\right)-14y=504$$

Multiply 45 times -\frac{2y}{3}+20.

$$-30y+900-14y=504$$

Add -30y to -14y.

$$-44y+900=504$$

Subtract 900 from both sides of the equation.

$$-44y=-396$$

Divide both sides by -44.

$$y=9$$

Substitute 9 for y in x=-\frac{2}{3}y+20. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\times 9+20$$

Multiply -\frac{2}{3} times 9.

$$x=-6+20$$

Add 20 to -6.

$$x=14$$

The system is now solved.

$$x=14,y=9$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x+2y=60$$

Consider the second equation. Multiply both sides of the equation by 63, the least common multiple of 7,9.

$$9\times 5x-7\times 2y=504$$

Multiply 9 and 5 to get 45.

$$45x-7\times 2y=504$$

Multiply -7 and 2 to get -14.

$$45x-14y=504$$

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

$$3x+2y=60,45x-14y=504$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&2\\45&-14\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}60\\504\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\45&-14\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&2\\45&-14\end{matrix}\right))\left(\begin{matrix}3&2\\45&-14\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\45&-14\end{matrix}\right))\left(\begin{matrix}60\\504\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\45&-14\end{matrix}\right))\left(\begin{matrix}60\\504\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\45&-14\end{matrix}\right))\left(\begin{matrix}60\\504\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{14}{3\left(-14\right)-2\times 45}&-\frac{2}{3\left(-14\right)-2\times 45}\\-\frac{45}{3\left(-14\right)-2\times 45}&\frac{3}{3\left(-14\right)-2\times 45}\end{matrix}\right)\left(\begin{matrix}60\\504\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{66}&\frac{1}{66}\\\frac{15}{44}&-\frac{1}{44}\end{matrix}\right)\left(\begin{matrix}60\\504\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{66}\times 60+\frac{1}{66}\times 504\\\frac{15}{44}\times 60-\frac{1}{44}\times 504\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\9\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=14,y=9$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3x+2y=60$$

Consider the second equation. Multiply both sides of the equation by 63, the least common multiple of 7,9.

$$9\times 5x-7\times 2y=504$$

Multiply 9 and 5 to get 45.

$$45x-7\times 2y=504$$

Multiply -7 and 2 to get -14.

$$45x-14y=504$$

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.

$$3x+2y=60,45x-14y=504$$

To make 3x and 45x equal, multiply all terms on each side of the first equation by 45 and all terms on each side of the second by 3.

$$45\times 3x+45\times 2y=45\times 60,3\times 45x+3\left(-14\right)y=3\times 504$$

Simplify.

$$135x+90y=2700,135x-42y=1512$$

Subtract 135x-42y=1512 from 135x+90y=2700 by subtracting like terms on each side of the equal sign.

$$135x-135x+90y+42y=2700-1512$$

Add 135x to -135x. Terms 135x and -135x cancel out, leaving an equation with only one variable that can be solved.

$$90y+42y=2700-1512$$

Add 90y to 42y.

$$132y=2700-1512$$

Add 2700 to -1512.

$$132y=1188$$

Divide both sides by 132.

$$y=9$$

Substitute 9 for y in 45x-14y=504. Because the resulting equation contains only one variable, you can solve for x directly.

$$45x-14\times 9=504$$

Multiply -14 times 9.

$$45x-126=504$$

Add 126 to both sides of the equation.

$$45x=630$$

Divide both sides by 45.

$$x=14$$

The system is now solved.

$$x=14,y=9$$