Solve {l}{x+1+2(y+4)=10}{1+x/2+2=y-1/3-2x}

$$\left\{ \begin{array} { l } { x + 1 + 2 ( y + 4 ) = 10 } \\ { \frac { 1 + x } { 2 } + 2 = \frac { y - 1 } { 3 } - 2 x } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=1$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 2 by y+4.

$$x+1+2y+8=10$$

Add 1 and 8 to get 9.

$$x+9+2y=10$$

Subtract 9 from both sides.

$$x+2y=10-9$$

Subtract 9 from 10 to get 1.

$$x+2y=1$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(1+x\right)+12=2\left(y-1\right)-12x$$

Use the distributive property to multiply 3 by 1+x.

$$3+3x+12=2\left(y-1\right)-12x$$

Add 3 and 12 to get 15.

$$15+3x=2\left(y-1\right)-12x$$

Use the distributive property to multiply 2 by y-1.

$$15+3x=2y-2-12x$$

Subtract 2y from both sides.

$$15+3x-2y=-2-12x$$

Add 12x to both sides.

$$15+3x-2y+12x=-2$$

Combine 3x and 12x to get 15x.

$$15+15x-2y=-2$$

Subtract 15 from both sides.

$$15x-2y=-2-15$$

Subtract 15 from -2 to get -17.

$$15x-2y=-17$$

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.

$$x+2y=1,15x-2y=-17$$

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

$$x+2y=1$$

Subtract 2y from both sides of the equation.

$$x=-2y+1$$

Substitute -2y+1 for x in the other equation, 15x-2y=-17.

$$15\left(-2y+1\right)-2y=-17$$

Multiply 15 times -2y+1.

$$-30y+15-2y=-17$$

Add -30y to -2y.

$$-32y+15=-17$$

Subtract 15 from both sides of the equation.

$$-32y=-32$$

Divide both sides by -32.

$$y=1$$

Substitute 1 for y in x=-2y+1. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2+1$$

Add 1 to -2.

$$x=-1$$

The system is now solved.

$$x=-1,y=1$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 2 by y+4.

$$x+1+2y+8=10$$

Add 1 and 8 to get 9.

$$x+9+2y=10$$

Subtract 9 from both sides.

$$x+2y=10-9$$

Subtract 9 from 10 to get 1.

$$x+2y=1$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(1+x\right)+12=2\left(y-1\right)-12x$$

Use the distributive property to multiply 3 by 1+x.

$$3+3x+12=2\left(y-1\right)-12x$$

Add 3 and 12 to get 15.

$$15+3x=2\left(y-1\right)-12x$$

Use the distributive property to multiply 2 by y-1.

$$15+3x=2y-2-12x$$

Subtract 2y from both sides.

$$15+3x-2y=-2-12x$$

Add 12x to both sides.

$$15+3x-2y+12x=-2$$

Combine 3x and 12x to get 15x.

$$15+15x-2y=-2$$

Subtract 15 from both sides.

$$15x-2y=-2-15$$

Subtract 15 from -2 to get -17.

$$15x-2y=-17$$

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

$$x+2y=1,15x-2y=-17$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&2\\15&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\-17\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\15&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&2\\15&-2\end{matrix}\right))\left(\begin{matrix}1&2\\15&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\15&-2\end{matrix}\right))\left(\begin{matrix}1\\-17\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\15&-2\end{matrix}\right))\left(\begin{matrix}1\\-17\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\15&-2\end{matrix}\right))\left(\begin{matrix}1\\-17\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2-2\times 15}&-\frac{2}{-2-2\times 15}\\-\frac{15}{-2-2\times 15}&\frac{1}{-2-2\times 15}\end{matrix}\right)\left(\begin{matrix}1\\-17\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{16}&\frac{1}{16}\\\frac{15}{32}&-\frac{1}{32}\end{matrix}\right)\left(\begin{matrix}1\\-17\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{16}+\frac{1}{16}\left(-17\right)\\\frac{15}{32}-\frac{1}{32}\left(-17\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-1,y=1$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 2 by y+4.

$$x+1+2y+8=10$$

Add 1 and 8 to get 9.

$$x+9+2y=10$$

Subtract 9 from both sides.

$$x+2y=10-9$$

Subtract 9 from 10 to get 1.

$$x+2y=1$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.

$$3\left(1+x\right)+12=2\left(y-1\right)-12x$$

Use the distributive property to multiply 3 by 1+x.

$$3+3x+12=2\left(y-1\right)-12x$$

Add 3 and 12 to get 15.

$$15+3x=2\left(y-1\right)-12x$$

Use the distributive property to multiply 2 by y-1.

$$15+3x=2y-2-12x$$

Subtract 2y from both sides.

$$15+3x-2y=-2-12x$$

Add 12x to both sides.

$$15+3x-2y+12x=-2$$

Combine 3x and 12x to get 15x.

$$15+15x-2y=-2$$

Subtract 15 from both sides.

$$15x-2y=-2-15$$

Subtract 15 from -2 to get -17.

$$15x-2y=-17$$

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.

$$x+2y=1,15x-2y=-17$$

To make x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 1.

$$15x+15\times 2y=15,15x-2y=-17$$

Simplify.

$$15x+30y=15,15x-2y=-17$$

Subtract 15x-2y=-17 from 15x+30y=15 by subtracting like terms on each side of the equal sign.

$$15x-15x+30y+2y=15+17$$

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

$$30y+2y=15+17$$

Add 30y to 2y.

$$32y=15+17$$

Add 15 to 17.

$$32y=32$$

Divide both sides by 32.

$$y=1$$

Substitute 1 for y in 15x-2y=-17. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x-2=-17$$

Add 2 to both sides of the equation.

$$15x=-15$$

Divide both sides by 15.

$$x=-1$$

The system is now solved.

$$x=-1,y=1$$