Solve 302x:y=7:5quadtext{e}quadx+y=24

$$302 x : y = 7 : 5 \quad \text { e } \quad x + y = 24$$

Solve for x, y

$$x=\frac{168}{1517}\approx 0.110744891$$

$$y = \frac{36240}{1517} = 23\frac{1349}{1517} \approx 23.889255109$$

Steps Using Substitution

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5y, the least common multiple of y,5.

$$5\times 302x=7y$$

Multiply 5 and 302 to get 1510.

$$1510x=7y$$

Divide both sides by 1510.

$$x=\frac{1}{1510}\times 7y$$

Multiply \frac{1}{1510} times 7y.

$$x=\frac{7}{1510}y$$

Substitute \frac{7y}{1510} for x in the other equation, x+y=24.

$$\frac{7}{1510}y+y=24$$

Add \frac{7y}{1510} to y.

$$\frac{1517}{1510}y=24$$

Divide both sides of the equation by \frac{1517}{1510}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{36240}{1517}$$

Substitute \frac{36240}{1517} for y in x=\frac{7}{1510}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{7}{1510}\times \frac{36240}{1517}$$

Multiply \frac{7}{1510} times \frac{36240}{1517} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{168}{1517}$$

The system is now solved.

$$x=\frac{168}{1517},y=\frac{36240}{1517}$$

Steps Using Matrices

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5y, the least common multiple of y,5.

$$5\times 302x=7y$$

Multiply 5 and 302 to get 1510.

$$1510x=7y$$

Subtract 7y from both sides.

$$1510x-7y=0$$

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

$$1510x-7y=0,x+y=24$$

Write the equations in matrix form.

$$\left(\begin{matrix}1510&-7\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\24\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1510&-7\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1510&-7\\1&1\end{matrix}\right))\left(\begin{matrix}1510&-7\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1510&-7\\1&1\end{matrix}\right))\left(\begin{matrix}0\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1510&-7\\1&1\end{matrix}\right))\left(\begin{matrix}0\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1510&-7\\1&1\end{matrix}\right))\left(\begin{matrix}0\\24\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1510-\left(-7\right)}&-\frac{-7}{1510-\left(-7\right)}\\-\frac{1}{1510-\left(-7\right)}&\frac{1510}{1510-\left(-7\right)}\end{matrix}\right)\left(\begin{matrix}0\\24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1517}&\frac{7}{1517}\\-\frac{1}{1517}&\frac{1510}{1517}\end{matrix}\right)\left(\begin{matrix}0\\24\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{1517}\times 24\\\frac{1510}{1517}\times 24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{168}{1517}\\\frac{36240}{1517}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{168}{1517},y=\frac{36240}{1517}$$

Steps Using Elimination

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 5y, the least common multiple of y,5.

$$5\times 302x=7y$$

Multiply 5 and 302 to get 1510.

$$1510x=7y$$

Subtract 7y from both sides.

$$1510x-7y=0$$

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.

$$1510x-7y=0,x+y=24$$

To make 1510x 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 1510.

$$1510x-7y=0,1510x+1510y=1510\times 24$$

Simplify.

$$1510x-7y=0,1510x+1510y=36240$$

Subtract 1510x+1510y=36240 from 1510x-7y=0 by subtracting like terms on each side of the equal sign.

$$1510x-1510x-7y-1510y=-36240$$

Add 1510x to -1510x. Terms 1510x and -1510x cancel out, leaving an equation with only one variable that can be solved.

$$-7y-1510y=-36240$$

Add -7y to -1510y.

$$-1517y=-36240$$

Divide both sides by -1517.

$$y=\frac{36240}{1517}$$

Substitute \frac{36240}{1517} for y in x+y=24. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+\frac{36240}{1517}=24$$

Subtract \frac{36240}{1517} from both sides of the equation.

$$x=\frac{168}{1517}$$

The system is now solved.

$$x=\frac{168}{1517},y=\frac{36240}{1517}$$