Solve {l}{x+1=5(y+2)}{3(2x-5)=5+4(3y+1)}

$$\left\{ \begin{array} { l } { x + 1 = 5 ( y + 2 ) } \\ { 3 ( 2 x - 5 ) = 5 + 4 ( 3 y + 1 ) } \end{array} \right.$$

Solve for x, y

$$x=\frac{2}{3}\approx 0.666666667$$

$$y = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 5 by y+2.

$$x+1=5y+10$$

Subtract 5y from both sides.

$$x+1-5y=10$$

Subtract 1 from both sides.

$$x-5y=10-1$$

Subtract 1 from 10 to get 9.

$$x-5y=9$$

Consider the second equation. Use the distributive property to multiply 3 by 2x-5.

$$6x-15=5+4\left(3y+1\right)$$

Use the distributive property to multiply 4 by 3y+1.

$$6x-15=5+12y+4$$

Add 5 and 4 to get 9.

$$6x-15=9+12y$$

Subtract 12y from both sides.

$$6x-15-12y=9$$

Add 15 to both sides.

$$6x-12y=9+15$$

Add 9 and 15 to get 24.

$$6x-12y=24$$

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.

$$x-5y=9,6x-12y=24$$

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

$$x-5y=9$$

Add 5y to both sides of the equation.

$$x=5y+9$$

Substitute 5y+9 for x in the other equation, 6x-12y=24.

$$6\left(5y+9\right)-12y=24$$

Multiply 6 times 5y+9.

$$30y+54-12y=24$$

Add 30y to -12y.

$$18y+54=24$$

Subtract 54 from both sides of the equation.

$$18y=-30$$

Divide both sides by 18.

$$y=-\frac{5}{3}$$

Substitute -\frac{5}{3} for y in x=5y+9. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=5\left(-\frac{5}{3}\right)+9$$

Multiply 5 times -\frac{5}{3}.

$$x=-\frac{25}{3}+9$$

Add 9 to -\frac{25}{3}.

$$x=\frac{2}{3}$$

The system is now solved.

$$x=\frac{2}{3},y=-\frac{5}{3}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 5 by y+2.

$$x+1=5y+10$$

Subtract 5y from both sides.

$$x+1-5y=10$$

Subtract 1 from both sides.

$$x-5y=10-1$$

Subtract 1 from 10 to get 9.

$$x-5y=9$$

Consider the second equation. Use the distributive property to multiply 3 by 2x-5.

$$6x-15=5+4\left(3y+1\right)$$

Use the distributive property to multiply 4 by 3y+1.

$$6x-15=5+12y+4$$

Add 5 and 4 to get 9.

$$6x-15=9+12y$$

Subtract 12y from both sides.

$$6x-15-12y=9$$

Add 15 to both sides.

$$6x-12y=9+15$$

Add 9 and 15 to get 24.

$$6x-12y=24$$

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

$$x-5y=9,6x-12y=24$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-5\\6&-12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9\\24\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-5\\6&-12\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-5\\6&-12\end{matrix}\right))\left(\begin{matrix}1&-5\\6&-12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-5\\6&-12\end{matrix}\right))\left(\begin{matrix}9\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-5\\6&-12\end{matrix}\right))\left(\begin{matrix}9\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-5\\6&-12\end{matrix}\right))\left(\begin{matrix}9\\24\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{12}{-12-\left(-5\times 6\right)}&-\frac{-5}{-12-\left(-5\times 6\right)}\\-\frac{6}{-12-\left(-5\times 6\right)}&\frac{1}{-12-\left(-5\times 6\right)}\end{matrix}\right)\left(\begin{matrix}9\\24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3}&\frac{5}{18}\\-\frac{1}{3}&\frac{1}{18}\end{matrix}\right)\left(\begin{matrix}9\\24\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3}\times 9+\frac{5}{18}\times 24\\-\frac{1}{3}\times 9+\frac{1}{18}\times 24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\\-\frac{5}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{2}{3},y=-\frac{5}{3}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 5 by y+2.

$$x+1=5y+10$$

Subtract 5y from both sides.

$$x+1-5y=10$$

Subtract 1 from both sides.

$$x-5y=10-1$$

Subtract 1 from 10 to get 9.

$$x-5y=9$$

Consider the second equation. Use the distributive property to multiply 3 by 2x-5.

$$6x-15=5+4\left(3y+1\right)$$

Use the distributive property to multiply 4 by 3y+1.

$$6x-15=5+12y+4$$

Add 5 and 4 to get 9.

$$6x-15=9+12y$$

Subtract 12y from both sides.

$$6x-15-12y=9$$

Add 15 to both sides.

$$6x-12y=9+15$$

Add 9 and 15 to get 24.

$$6x-12y=24$$

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.

$$x-5y=9,6x-12y=24$$

To make x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 1.

$$6x+6\left(-5\right)y=6\times 9,6x-12y=24$$

Simplify.

$$6x-30y=54,6x-12y=24$$

Subtract 6x-12y=24 from 6x-30y=54 by subtracting like terms on each side of the equal sign.

$$6x-6x-30y+12y=54-24$$

Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.

$$-30y+12y=54-24$$

Add -30y to 12y.

$$-18y=54-24$$

Add 54 to -24.

$$-18y=30$$

Divide both sides by -18.

$$y=-\frac{5}{3}$$

Substitute -\frac{5}{3} for y in 6x-12y=24. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x-12\left(-\frac{5}{3}\right)=24$$

Multiply -12 times -\frac{5}{3}.

$$6x+20=24$$

Subtract 20 from both sides of the equation.

$$6x=4$$

Divide both sides by 6.

$$x=\frac{2}{3}$$

The system is now solved.

$$x=\frac{2}{3},y=-\frac{5}{3}$$