Solve {l}{y/x=7/5}{y-3/x-9=4/3}

$$\left. \begin{array} { l } { \frac { y } { x } = \frac { 7 } { 5 } } \\ { \frac { y - 3 } { x - 9 } = \frac { 4 } { 3 } } \end{array} \right.$$

Solve for y, x

$$x=-135$$

$$y=-189$$

Steps Using Substitution

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

$$5y=7x$$

Divide both sides by 5.

$$y=\frac{1}{5}\times 7x$$

Multiply \frac{1}{5} times 7x.

$$y=\frac{7}{5}x$$

Substitute \frac{7x}{5} for y in the other equation, 3y-4x=-27.

$$3\times \frac{7}{5}x-4x=-27$$

Multiply 3 times \frac{7x}{5}.

$$\frac{21}{5}x-4x=-27$$

Add \frac{21x}{5} to -4x.

$$\frac{1}{5}x=-27$$

Multiply both sides by 5.

$$x=-135$$

Substitute -135 for x in y=\frac{7}{5}x. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{7}{5}\left(-135\right)$$

Multiply \frac{7}{5} times -135.

$$y=-189$$

The system is now solved.

$$y=-189,x=-135$$

Steps Using Matrices

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

$$5y=7x$$

Subtract 7x from both sides.

$$5y-7x=0$$

Consider the second equation. Variable x cannot be equal to 9 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-9\right), the least common multiple of x-9,3.

$$3\left(y-3\right)=4\left(x-9\right)$$

Use the distributive property to multiply 3 by y-3.

$$3y-9=4\left(x-9\right)$$

Use the distributive property to multiply 4 by x-9.

$$3y-9=4x-36$$

Subtract 4x from both sides.

$$3y-9-4x=-36$$

Add 9 to both sides.

$$3y-4x=-36+9$$

Add -36 and 9 to get -27.

$$3y-4x=-27$$

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

$$5y-7x=0,3y-4x=-27$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-7\\3&-4\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\-27\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-7\\3&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-7\\3&-4\end{matrix}\right))\left(\begin{matrix}5&-7\\3&-4\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}5&-7\\3&-4\end{matrix}\right))\left(\begin{matrix}0\\-27\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}5&-7\\3&-4\end{matrix}\right))\left(\begin{matrix}0\\-27\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}5&-7\\3&-4\end{matrix}\right))\left(\begin{matrix}0\\-27\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5\left(-4\right)-\left(-7\times 3\right)}&-\frac{-7}{5\left(-4\right)-\left(-7\times 3\right)}\\-\frac{3}{5\left(-4\right)-\left(-7\times 3\right)}&\frac{5}{5\left(-4\right)-\left(-7\times 3\right)}\end{matrix}\right)\left(\begin{matrix}0\\-27\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-4&7\\-3&5\end{matrix}\right)\left(\begin{matrix}0\\-27\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}7\left(-27\right)\\5\left(-27\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-189\\-135\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-189,x=-135$$

Steps Using Elimination

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

$$5y=7x$$

Subtract 7x from both sides.

$$5y-7x=0$$

Consider the second equation. Variable x cannot be equal to 9 since division by zero is not defined. Multiply both sides of the equation by 3\left(x-9\right), the least common multiple of x-9,3.

$$3\left(y-3\right)=4\left(x-9\right)$$

Use the distributive property to multiply 3 by y-3.

$$3y-9=4\left(x-9\right)$$

Use the distributive property to multiply 4 by x-9.

$$3y-9=4x-36$$

Subtract 4x from both sides.

$$3y-9-4x=-36$$

Add 9 to both sides.

$$3y-4x=-36+9$$

Add -36 and 9 to get -27.

$$3y-4x=-27$$

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.

$$5y-7x=0,3y-4x=-27$$

To make 5y and 3y equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 5.

$$3\times 5y+3\left(-7\right)x=0,5\times 3y+5\left(-4\right)x=5\left(-27\right)$$

Simplify.

$$15y-21x=0,15y-20x=-135$$

Subtract 15y-20x=-135 from 15y-21x=0 by subtracting like terms on each side of the equal sign.

$$15y-15y-21x+20x=135$$

Add 15y to -15y. Terms 15y and -15y cancel out, leaving an equation with only one variable that can be solved.

$$-21x+20x=135$$

Add -21x to 20x.

$$-x=135$$

Divide both sides by -1.

$$x=-135$$

Substitute -135 for x in 3y-4x=-27. Because the resulting equation contains only one variable, you can solve for y directly.

$$3y-4\left(-135\right)=-27$$

Multiply -4 times -135.

$$3y+540=-27$$

Subtract 540 from both sides of the equation.

$$3y=-567$$

Divide both sides by 3.

$$y=-189$$

The system is now solved.

$$y=-189,x=-135$$