Solve x/5=y/4quadtext{u}quadx+y=63

$$\frac { x } { 5 } = \frac { y } { 4 } \quad \text { u } \quad x + y = 63$$

Solve for x, y

$$x=35$$

$$y=28$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4x=5y$$

Divide both sides by 4.

$$x=\frac{1}{4}\times 5y$$

Multiply \frac{1}{4} times 5y.

$$x=\frac{5}{4}y$$

Substitute \frac{5y}{4} for x in the other equation, x+y=63.

$$\frac{5}{4}y+y=63$$

Add \frac{5y}{4} to y.

$$\frac{9}{4}y=63$$

Divide both sides of the equation by \frac{9}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=28$$

Substitute 28 for y in x=\frac{5}{4}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{4}\times 28$$

Multiply \frac{5}{4} times 28.

$$x=35$$

The system is now solved.

$$x=35,y=28$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4x=5y$$

Subtract 5y from both sides.

$$4x-5y=0$$

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

$$4x-5y=0,x+y=63$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&-5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\63\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&-5\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&-5\\1&1\end{matrix}\right))\left(\begin{matrix}4&-5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-5\\1&1\end{matrix}\right))\left(\begin{matrix}0\\63\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-5\\1&1\end{matrix}\right))\left(\begin{matrix}0\\63\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-5\\1&1\end{matrix}\right))\left(\begin{matrix}0\\63\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4-\left(-5\right)}&-\frac{-5}{4-\left(-5\right)}\\-\frac{1}{4-\left(-5\right)}&\frac{4}{4-\left(-5\right)}\end{matrix}\right)\left(\begin{matrix}0\\63\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}&\frac{5}{9}\\-\frac{1}{9}&\frac{4}{9}\end{matrix}\right)\left(\begin{matrix}0\\63\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{9}\times 63\\\frac{4}{9}\times 63\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35\\28\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=35,y=28$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 20, the least common multiple of 5,4.

$$4x=5y$$

Subtract 5y from both sides.

$$4x-5y=0$$

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.

$$4x-5y=0,x+y=63$$

To make 4x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 4.

$$4x-5y=0,4x+4y=4\times 63$$

Simplify.

$$4x-5y=0,4x+4y=252$$

Subtract 4x+4y=252 from 4x-5y=0 by subtracting like terms on each side of the equal sign.

$$4x-4x-5y-4y=-252$$

Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.

$$-5y-4y=-252$$

Add -5y to -4y.

$$-9y=-252$$

Divide both sides by -9.

$$y=28$$

Substitute 28 for y in x+y=63. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+28=63$$

Subtract 28 from both sides of the equation.

$$x=35$$

The system is now solved.

$$x=35,y=28$$