Solve {l}{x+2/y+5=x+7/y}{2x-4y=-1}

$$\left\{ \begin{array} { l } { \frac { x + 2 } { y + 5 } = \frac { x + 7 } { y } } \\ { 2 x - 4 y = - 1 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{29}{6} = -4\frac{5}{6} \approx -4.833333333$$

$$y = -\frac{13}{6} = -2\frac{1}{6} \approx -2.166666667$$

Steps Using Substitution

Consider the first equation. Variable y cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by y\left(y+5\right), the least common multiple of y+5,y.

$$y\left(x+2\right)=\left(y+5\right)\left(x+7\right)$$

Use the distributive property to multiply y by x+2.

$$yx+2y=\left(y+5\right)\left(x+7\right)$$

Use the distributive property to multiply y+5 by x+7.

$$yx+2y=yx+7y+5x+35$$

Subtract yx from both sides.

$$yx+2y-yx=7y+5x+35$$

Combine yx and -yx to get 0.

$$2y=7y+5x+35$$

Subtract 7y from both sides.

$$2y-7y=5x+35$$

Combine 2y and -7y to get -5y.

$$-5y=5x+35$$

Divide both sides by -5.

$$y=-\frac{1}{5}\left(5x+35\right)$$

Multiply -\frac{1}{5} times 35+5x.

$$y=-x-7$$

Substitute -x-7 for y in the other equation, -4y+2x=-1.

$$-4\left(-x-7\right)+2x=-1$$

Multiply -4 times -x-7.

$$4x+28+2x=-1$$

Add 4x to 2x.

$$6x+28=-1$$

Subtract 28 from both sides of the equation.

$$6x=-29$$

Divide both sides by 6.

$$x=-\frac{29}{6}$$

Substitute -\frac{29}{6} for x in y=-x-7. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\left(-\frac{29}{6}\right)-7$$

Multiply -1 times -\frac{29}{6}.

$$y=\frac{29}{6}-7$$

Add -7 to \frac{29}{6}.

$$y=-\frac{13}{6}$$

The system is now solved.

$$y=-\frac{13}{6},x=-\frac{29}{6}$$

Steps Using Matrices

Consider the first equation. Variable y cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by y\left(y+5\right), the least common multiple of y+5,y.

$$y\left(x+2\right)=\left(y+5\right)\left(x+7\right)$$

Use the distributive property to multiply y by x+2.

$$yx+2y=\left(y+5\right)\left(x+7\right)$$

Use the distributive property to multiply y+5 by x+7.

$$yx+2y=yx+7y+5x+35$$

Subtract yx from both sides.

$$yx+2y-yx=7y+5x+35$$

Combine yx and -yx to get 0.

$$2y=7y+5x+35$$

Subtract 7y from both sides.

$$2y-7y=5x+35$$

Combine 2y and -7y to get -5y.

$$-5y=5x+35$$

Subtract 5x from both sides.

$$-5y-5x=35$$

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

$$-5y-5x=35,-4y+2x=-1$$

Write the equations in matrix form.

$$\left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}35\\-1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right))\left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right))\left(\begin{matrix}35\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right))\left(\begin{matrix}35\\-1\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-5\\-4&2\end{matrix}\right))\left(\begin{matrix}35\\-1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{-5\times 2-\left(-5\left(-4\right)\right)}&-\frac{-5}{-5\times 2-\left(-5\left(-4\right)\right)}\\-\frac{-4}{-5\times 2-\left(-5\left(-4\right)\right)}&-\frac{5}{-5\times 2-\left(-5\left(-4\right)\right)}\end{matrix}\right)\left(\begin{matrix}35\\-1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{15}&-\frac{1}{6}\\-\frac{2}{15}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}35\\-1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{15}\times 35-\frac{1}{6}\left(-1\right)\\-\frac{2}{15}\times 35+\frac{1}{6}\left(-1\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{6}\\-\frac{29}{6}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=-\frac{13}{6},x=-\frac{29}{6}$$

Steps Using Elimination

Consider the first equation. Variable y cannot be equal to any of the values -5,0 since division by zero is not defined. Multiply both sides of the equation by y\left(y+5\right), the least common multiple of y+5,y.

$$y\left(x+2\right)=\left(y+5\right)\left(x+7\right)$$

Use the distributive property to multiply y by x+2.

$$yx+2y=\left(y+5\right)\left(x+7\right)$$

Use the distributive property to multiply y+5 by x+7.

$$yx+2y=yx+7y+5x+35$$

Subtract yx from both sides.

$$yx+2y-yx=7y+5x+35$$

Combine yx and -yx to get 0.

$$2y=7y+5x+35$$

Subtract 7y from both sides.

$$2y-7y=5x+35$$

Combine 2y and -7y to get -5y.

$$-5y=5x+35$$

Subtract 5x from both sides.

$$-5y-5x=35$$

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.

$$-5y-5x=35,-4y+2x=-1$$

To make -5y and -4y equal, multiply all terms on each side of the first equation by -4 and all terms on each side of the second by -5.

$$-4\left(-5\right)y-4\left(-5\right)x=-4\times 35,-5\left(-4\right)y-5\times 2x=-5\left(-1\right)$$

Simplify.

$$20y+20x=-140,20y-10x=5$$

Subtract 20y-10x=5 from 20y+20x=-140 by subtracting like terms on each side of the equal sign.

$$20y-20y+20x+10x=-140-5$$

Add 20y to -20y. Terms 20y and -20y cancel out, leaving an equation with only one variable that can be solved.

$$20x+10x=-140-5$$

Add 20x to 10x.

$$30x=-140-5$$

Add -140 to -5.

$$30x=-145$$

Divide both sides by 30.

$$x=-\frac{29}{6}$$

Substitute -\frac{29}{6} for x in -4y+2x=-1. Because the resulting equation contains only one variable, you can solve for y directly.

$$-4y+2\left(-\frac{29}{6}\right)=-1$$

Multiply 2 times -\frac{29}{6}.

$$-4y-\frac{29}{3}=-1$$

Add \frac{29}{3} to both sides of the equation.

$$-4y=\frac{26}{3}$$

Divide both sides by -4.

$$y=-\frac{13}{6}$$

The system is now solved.

$$y=-\frac{13}{6},x=-\frac{29}{6}$$