Solve {l}{2x+4/3+4y+2/4=10/12}{x+5/5=frac{y-1}{2}}

$$\left\{ \begin{array} { l } { \frac { 2 x + 4 } { 3 } + \frac { 4 y + 2 } { 4 } = \frac { 10 } { 12 } } \\ { \frac { x + 5 } { 5 } = \frac { y - 1 } { 2 } } \end{array} \right.$$

Solve for x, y

$$x = -\frac{15}{4} = -3\frac{3}{4} = -3.75$$

$$y = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 3,4,12.

$$4\left(2x+4\right)+3\left(4y+2\right)=10$$

Use the distributive property to multiply 4 by 2x+4.

$$8x+16+3\left(4y+2\right)=10$$

Use the distributive property to multiply 3 by 4y+2.

$$8x+16+12y+6=10$$

Add 16 and 6 to get 22.

$$8x+22+12y=10$$

Subtract 22 from both sides.

$$8x+12y=10-22$$

Subtract 22 from 10 to get -12.

$$8x+12y=-12$$

Consider the second equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.

$$2\left(x+5\right)=5\left(y-1\right)$$

Use the distributive property to multiply 2 by x+5.

$$2x+10=5\left(y-1\right)$$

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

$$2x+10=5y-5$$

Subtract 5y from both sides.

$$2x+10-5y=-5$$

Subtract 10 from both sides.

$$2x-5y=-5-10$$

Subtract 10 from -5 to get -15.

$$2x-5y=-15$$

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.

$$8x+12y=-12,2x-5y=-15$$

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

$$8x+12y=-12$$

Subtract 12y from both sides of the equation.

$$8x=-12y-12$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-12y-12\right)$$

Multiply \frac{1}{8} times -12y-12.

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

Substitute \frac{-3y-3}{2} for x in the other equation, 2x-5y=-15.

$$2\left(-\frac{3}{2}y-\frac{3}{2}\right)-5y=-15$$

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

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

Add -3y to -5y.

$$-8y-3=-15$$

Add 3 to both sides of the equation.

$$-8y=-12$$

Divide both sides by -8.

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

Substitute \frac{3}{2} for y in x=-\frac{3}{2}y-\frac{3}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

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

Multiply -\frac{3}{2} times \frac{3}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{9}{4}-\frac{3}{2}$$

Add -\frac{3}{2} to -\frac{9}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{15}{4}$$

The system is now solved.

$$x=-\frac{15}{4},y=\frac{3}{2}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 3,4,12.

$$4\left(2x+4\right)+3\left(4y+2\right)=10$$

Use the distributive property to multiply 4 by 2x+4.

$$8x+16+3\left(4y+2\right)=10$$

Use the distributive property to multiply 3 by 4y+2.

$$8x+16+12y+6=10$$

Add 16 and 6 to get 22.

$$8x+22+12y=10$$

Subtract 22 from both sides.

$$8x+12y=10-22$$

Subtract 22 from 10 to get -12.

$$8x+12y=-12$$

Consider the second equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.

$$2\left(x+5\right)=5\left(y-1\right)$$

Use the distributive property to multiply 2 by x+5.

$$2x+10=5\left(y-1\right)$$

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

$$2x+10=5y-5$$

Subtract 5y from both sides.

$$2x+10-5y=-5$$

Subtract 10 from both sides.

$$2x-5y=-5-10$$

Subtract 10 from -5 to get -15.

$$2x-5y=-15$$

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

$$8x+12y=-12,2x-5y=-15$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&12\\2&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-12\\-15\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&12\\2&-5\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&12\\2&-5\end{matrix}\right))\left(\begin{matrix}8&12\\2&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&12\\2&-5\end{matrix}\right))\left(\begin{matrix}-12\\-15\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&12\\2&-5\end{matrix}\right))\left(\begin{matrix}-12\\-15\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&12\\2&-5\end{matrix}\right))\left(\begin{matrix}-12\\-15\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{8\left(-5\right)-12\times 2}&-\frac{12}{8\left(-5\right)-12\times 2}\\-\frac{2}{8\left(-5\right)-12\times 2}&\frac{8}{8\left(-5\right)-12\times 2}\end{matrix}\right)\left(\begin{matrix}-12\\-15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{64}&\frac{3}{16}\\\frac{1}{32}&-\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}-12\\-15\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{64}\left(-12\right)+\frac{3}{16}\left(-15\right)\\\frac{1}{32}\left(-12\right)-\frac{1}{8}\left(-15\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{15}{4}\\\frac{3}{2}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{15}{4},y=\frac{3}{2}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 3,4,12.

$$4\left(2x+4\right)+3\left(4y+2\right)=10$$

Use the distributive property to multiply 4 by 2x+4.

$$8x+16+3\left(4y+2\right)=10$$

Use the distributive property to multiply 3 by 4y+2.

$$8x+16+12y+6=10$$

Add 16 and 6 to get 22.

$$8x+22+12y=10$$

Subtract 22 from both sides.

$$8x+12y=10-22$$

Subtract 22 from 10 to get -12.

$$8x+12y=-12$$

Consider the second equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.

$$2\left(x+5\right)=5\left(y-1\right)$$

Use the distributive property to multiply 2 by x+5.

$$2x+10=5\left(y-1\right)$$

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

$$2x+10=5y-5$$

Subtract 5y from both sides.

$$2x+10-5y=-5$$

Subtract 10 from both sides.

$$2x-5y=-5-10$$

Subtract 10 from -5 to get -15.

$$2x-5y=-15$$

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.

$$8x+12y=-12,2x-5y=-15$$

To make 8x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 8.

$$2\times 8x+2\times 12y=2\left(-12\right),8\times 2x+8\left(-5\right)y=8\left(-15\right)$$

Simplify.

$$16x+24y=-24,16x-40y=-120$$

Subtract 16x-40y=-120 from 16x+24y=-24 by subtracting like terms on each side of the equal sign.

$$16x-16x+24y+40y=-24+120$$

Add 16x to -16x. Terms 16x and -16x cancel out, leaving an equation with only one variable that can be solved.

$$24y+40y=-24+120$$

Add 24y to 40y.

$$64y=-24+120$$

Add -24 to 120.

$$64y=96$$

Divide both sides by 64.

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

Substitute \frac{3}{2} for y in 2x-5y=-15. Because the resulting equation contains only one variable, you can solve for x directly.

$$2x-5\times \frac{3}{2}=-15$$

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

$$2x-\frac{15}{2}=-15$$

Add \frac{15}{2} to both sides of the equation.

$$2x=-\frac{15}{2}$$

Divide both sides by 2.

$$x=-\frac{15}{4}$$

The system is now solved.

$$x=-\frac{15}{4},y=\frac{3}{2}$$