Solve 7x-4/y=10text{intheform}y=mx+b?

$$\frac { 7 x - 4 } { y } = 10 \text { in the form } y = m x + b ?$$

Solve for x, y

$$\left\{\begin{matrix}x=\frac{2\left(5b+2\right)}{7-10m}\text{, }y=\frac{4m+7b}{7-10m}\text{, }&m\neq -\frac{7b}{4}\text{ and }m\neq \frac{7}{10}\\x=\frac{2\left(5y+2\right)}{7}\text{, }y\neq 0\text{, }&m=\frac{7}{10}\text{ and }b=-\frac{2}{5}\end{matrix}\right.$$

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 y.

$$7x-4=10y$$

Subtract 10y from both sides.

$$7x-4-10y=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$7x-10y=4$$

Consider the second equation. Subtract mx from both sides.

$$y-mx=b$$

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.

$$7x-10y=4,\left(-m\right)x+y=b$$

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

$$7x-10y=4$$

Add 10y to both sides of the equation.

$$7x=10y+4$$

Divide both sides by 7.

$$x=\frac{1}{7}\left(10y+4\right)$$

Multiply \frac{1}{7} times 10y+4.

$$x=\frac{10}{7}y+\frac{4}{7}$$

Substitute \frac{10y+4}{7} for x in the other equation, \left(-m\right)x+y=b.

$$\left(-m\right)\left(\frac{10}{7}y+\frac{4}{7}\right)+y=b$$

Multiply -m times \frac{10y+4}{7}.

$$\left(-\frac{10m}{7}\right)y-\frac{4m}{7}+y=b$$

Add -\frac{10my}{7} to y.

$$\left(-\frac{10m}{7}+1\right)y-\frac{4m}{7}=b$$

Add \frac{4m}{7} to both sides of the equation.

$$\left(-\frac{10m}{7}+1\right)y=\frac{4m}{7}+b$$

Divide both sides by -\frac{10m}{7}+1.

$$y=\frac{4m+7b}{7-10m}$$

Substitute \frac{7b+4m}{-10m+7} for y in x=\frac{10}{7}y+\frac{4}{7}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{10}{7}\times \frac{4m+7b}{7-10m}+\frac{4}{7}$$

Multiply \frac{10}{7} times \frac{7b+4m}{-10m+7}.

$$x=\frac{10\left(4m+7b\right)}{7\left(7-10m\right)}+\frac{4}{7}$$

Add \frac{4}{7} to \frac{10\left(7b+4m\right)}{7\left(-10m+7\right)}.

$$x=\frac{2\left(5b+2\right)}{7-10m}$$

The system is now solved.

$$x=\frac{2\left(5b+2\right)}{7-10m},y=\frac{4m+7b}{7-10m}$$

Variable y cannot be equal to 0.

$$x=\frac{2\left(5b+2\right)}{7-10m},y=\frac{4m+7b}{7-10m}\text{, }y\neq 0$$

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 y.

$$7x-4=10y$$

Subtract 10y from both sides.

$$7x-4-10y=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$7x-10y=4$$

Consider the second equation. Subtract mx from both sides.

$$y-mx=b$$

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

$$7x-10y=4,\left(-m\right)x+y=b$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&-10\\-m&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\b\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-10\\-m&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&-10\\-m&1\end{matrix}\right))\left(\begin{matrix}7&-10\\-m&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-10\\-m&1\end{matrix}\right))\left(\begin{matrix}4\\b\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-10\\-m&1\end{matrix}\right))\left(\begin{matrix}4\\b\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-10\\-m&1\end{matrix}\right))\left(\begin{matrix}4\\b\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7-\left(-10\left(-m\right)\right)}&-\frac{-10}{7-\left(-10\left(-m\right)\right)}\\-\frac{-m}{7-\left(-10\left(-m\right)\right)}&\frac{7}{7-\left(-10\left(-m\right)\right)}\end{matrix}\right)\left(\begin{matrix}4\\b\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7-10m}&\frac{10}{7-10m}\\\frac{m}{7-10m}&\frac{7}{7-10m}\end{matrix}\right)\left(\begin{matrix}4\\b\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{7-10m}\times 4+\frac{10}{7-10m}b\\\frac{m}{7-10m}\times 4+\frac{7}{7-10m}b\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2\left(5b+2\right)}{10m-7}\\-\frac{4m+7b}{10m-7}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{2\left(5b+2\right)}{10m-7},y=-\frac{4m+7b}{10m-7}$$

Variable y cannot be equal to 0.

$$x=-\frac{2\left(5b+2\right)}{10m-7},y=-\frac{4m+7b}{10m-7}\text{, }y\neq 0$$

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 y.

$$7x-4=10y$$

Subtract 10y from both sides.

$$7x-4-10y=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$7x-10y=4$$

Consider the second equation. Subtract mx from both sides.

$$y-mx=b$$

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.

$$7x-10y=4,\left(-m\right)x+y=b$$

To make 7x and -mx equal, multiply all terms on each side of the first equation by -m and all terms on each side of the second by 7.

$$\left(-m\right)\times 7x+\left(-m\right)\left(-10\right)y=\left(-m\right)\times 4,7\left(-m\right)x+7y=7b$$

Simplify.

$$\left(-7m\right)x+10my=-4m,\left(-7m\right)x+7y=7b$$

Subtract \left(-7m\right)x+7y=7b from \left(-7m\right)x+10my=-4m by subtracting like terms on each side of the equal sign.

$$\left(-7m\right)x+7mx+10my-7y=-4m-7b$$

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

$$10my-7y=-4m-7b$$

Add 10my to -7y.

$$\left(10m-7\right)y=-4m-7b$$

Divide both sides by 10m-7.

$$y=-\frac{4m+7b}{10m-7}$$

Substitute -\frac{4m+7b}{10m-7} for y in \left(-m\right)x+y=b. Because the resulting equation contains only one variable, you can solve for x directly.

$$\left(-m\right)x-\frac{4m+7b}{10m-7}=b$$

Add \frac{4m+7b}{10m-7} to both sides of the equation.

$$\left(-m\right)x=\frac{2m\left(5b+2\right)}{10m-7}$$

Divide both sides by -m.

$$x=-\frac{2\left(5b+2\right)}{10m-7}$$

The system is now solved.

$$x=-\frac{2\left(5b+2\right)}{10m-7},y=-\frac{4m+7b}{10m-7}$$

Variable y cannot be equal to 0.

$$x=-\frac{2\left(5b+2\right)}{10m-7},y=-\frac{4m+7b}{10m-7}\text{, }y\neq 0$$