Solve {l}{5(x-1)=y+5}{5(y-1)=3(x+5)}

$$\left\{ \begin{array} { l } { 5 ( x - 1 ) = y + 5 } \\ { 5 ( y - 1 ) = 3 ( x + 5 ) } \end{array} \right.$$

Solve for x, y

$$x = \frac{35}{11} = 3\frac{2}{11} \approx 3.181818182$$

$$y = \frac{65}{11} = 5\frac{10}{11} \approx 5.909090909$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 5 by x-1.

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

Subtract y from both sides.

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

Add 5 to both sides.

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

Add 5 and 5 to get 10.

$$5x-y=10$$

Consider the second equation. Use the distributive property to multiply 5 by y-1.

$$5y-5=3\left(x+5\right)$$

Use the distributive property to multiply 3 by x+5.

$$5y-5=3x+15$$

Subtract 3x from both sides.

$$5y-5-3x=15$$

Add 5 to both sides.

$$5y-3x=15+5$$

Add 15 and 5 to get 20.

$$5y-3x=20$$

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.

$$5x-y=10,-3x+5y=20$$

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

$$5x-y=10$$

Add y to both sides of the equation.

$$5x=y+10$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(y+10\right)$$

Multiply \frac{1}{5} times y+10.

$$x=\frac{1}{5}y+2$$

Substitute \frac{y}{5}+2 for x in the other equation, -3x+5y=20.

$$-3\left(\frac{1}{5}y+2\right)+5y=20$$

Multiply -3 times \frac{y}{5}+2.

$$-\frac{3}{5}y-6+5y=20$$

Add -\frac{3y}{5} to 5y.

$$\frac{22}{5}y-6=20$$

Add 6 to both sides of the equation.

$$\frac{22}{5}y=26$$

Divide both sides of the equation by \frac{22}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{65}{11}$$

Substitute \frac{65}{11} for y in x=\frac{1}{5}y+2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{5}\times \frac{65}{11}+2$$

Multiply \frac{1}{5} times \frac{65}{11} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{13}{11}+2$$

Add 2 to \frac{13}{11}.

$$x=\frac{35}{11}$$

The system is now solved.

$$x=\frac{35}{11},y=\frac{65}{11}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 5 by x-1.

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

Subtract y from both sides.

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

Add 5 to both sides.

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

Add 5 and 5 to get 10.

$$5x-y=10$$

Consider the second equation. Use the distributive property to multiply 5 by y-1.

$$5y-5=3\left(x+5\right)$$

Use the distributive property to multiply 3 by x+5.

$$5y-5=3x+15$$

Subtract 3x from both sides.

$$5y-5-3x=15$$

Add 5 to both sides.

$$5y-3x=15+5$$

Add 15 and 5 to get 20.

$$5y-3x=20$$

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

$$5x-y=10,-3x+5y=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-1\\-3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-1\\-3&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-1\\-3&5\end{matrix}\right))\left(\begin{matrix}5&-1\\-3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-1\\-3&5\end{matrix}\right))\left(\begin{matrix}10\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-1\\-3&5\end{matrix}\right))\left(\begin{matrix}10\\20\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-1\\-3&5\end{matrix}\right))\left(\begin{matrix}10\\20\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5\times 5-\left(-\left(-3\right)\right)}&-\frac{-1}{5\times 5-\left(-\left(-3\right)\right)}\\-\frac{-3}{5\times 5-\left(-\left(-3\right)\right)}&\frac{5}{5\times 5-\left(-\left(-3\right)\right)}\end{matrix}\right)\left(\begin{matrix}10\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{22}&\frac{1}{22}\\\frac{3}{22}&\frac{5}{22}\end{matrix}\right)\left(\begin{matrix}10\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{22}\times 10+\frac{1}{22}\times 20\\\frac{3}{22}\times 10+\frac{5}{22}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{11}\\\frac{65}{11}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{35}{11},y=\frac{65}{11}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 5 by x-1.

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

Subtract y from both sides.

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

Add 5 to both sides.

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

Add 5 and 5 to get 10.

$$5x-y=10$$

Consider the second equation. Use the distributive property to multiply 5 by y-1.

$$5y-5=3\left(x+5\right)$$

Use the distributive property to multiply 3 by x+5.

$$5y-5=3x+15$$

Subtract 3x from both sides.

$$5y-5-3x=15$$

Add 5 to both sides.

$$5y-3x=15+5$$

Add 15 and 5 to get 20.

$$5y-3x=20$$

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.

$$5x-y=10,-3x+5y=20$$

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

$$-3\times 5x-3\left(-1\right)y=-3\times 10,5\left(-3\right)x+5\times 5y=5\times 20$$

Simplify.

$$-15x+3y=-30,-15x+25y=100$$

Subtract -15x+25y=100 from -15x+3y=-30 by subtracting like terms on each side of the equal sign.

$$-15x+15x+3y-25y=-30-100$$

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

$$3y-25y=-30-100$$

Add 3y to -25y.

$$-22y=-30-100$$

Add -30 to -100.

$$-22y=-130$$

Divide both sides by -22.

$$y=\frac{65}{11}$$

Substitute \frac{65}{11} for y in -3x+5y=20. Because the resulting equation contains only one variable, you can solve for x directly.

$$-3x+5\times \frac{65}{11}=20$$

Multiply 5 times \frac{65}{11}.

$$-3x+\frac{325}{11}=20$$

Subtract \frac{325}{11} from both sides of the equation.

$$-3x=-\frac{105}{11}$$

Divide both sides by -3.

$$x=\frac{35}{11}$$

The system is now solved.

$$x=\frac{35}{11},y=\frac{65}{11}$$