Solve {l}{3x-4y+3/4=4-4x-2y-9/3}{2x-y+3/3=4+x-2y+3/4}

$$\left. \begin{array} { l } { \frac { 3 x - 4 y + 3 } { 4 } = 4 - \frac { 4 x - 2 y - 9 } { 3 } } \\ { \frac { 2 x - y + 3 } { 3 } = 4 + \frac { x - 2 y + 3 } { 4 } } \end{array} \right.$$

Solve for x, y

$$x=7$$

$$y=5$$

Steps Using Substitution

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

$$3\left(3x-4y+3\right)=48-4\left(4x-2y-9\right)$$

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

$$9x-12y+9=48-4\left(4x-2y-9\right)$$

Use the distributive property to multiply -4 by 4x-2y-9.

$$9x-12y+9=48-16x+8y+36$$

Add 48 and 36 to get 84.

$$9x-12y+9=84-16x+8y$$

Add 16x to both sides.

$$9x-12y+9+16x=84+8y$$

Combine 9x and 16x to get 25x.

$$25x-12y+9=84+8y$$

Subtract 8y from both sides.

$$25x-12y+9-8y=84$$

Combine -12y and -8y to get -20y.

$$25x-20y+9=84$$

Subtract 9 from both sides.

$$25x-20y=84-9$$

Subtract 9 from 84 to get 75.

$$25x-20y=75$$

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

$$4\left(2x-y+3\right)=48+3\left(x-2y+3\right)$$

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

$$8x-4y+12=48+3\left(x-2y+3\right)$$

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

$$8x-4y+12=48+3x-6y+9$$

Add 48 and 9 to get 57.

$$8x-4y+12=57+3x-6y$$

Subtract 3x from both sides.

$$8x-4y+12-3x=57-6y$$

Combine 8x and -3x to get 5x.

$$5x-4y+12=57-6y$$

Add 6y to both sides.

$$5x-4y+12+6y=57$$

Combine -4y and 6y to get 2y.

$$5x+2y+12=57$$

Subtract 12 from both sides.

$$5x+2y=57-12$$

Subtract 12 from 57 to get 45.

$$5x+2y=45$$

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.

$$25x-20y=75,5x+2y=45$$

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

$$25x-20y=75$$

Add 20y to both sides of the equation.

$$25x=20y+75$$

Divide both sides by 25.

$$x=\frac{1}{25}\left(20y+75\right)$$

Multiply \frac{1}{25} times 20y+75.

$$x=\frac{4}{5}y+3$$

Substitute \frac{4y}{5}+3 for x in the other equation, 5x+2y=45.

$$5\left(\frac{4}{5}y+3\right)+2y=45$$

Multiply 5 times \frac{4y}{5}+3.

$$4y+15+2y=45$$

Add 4y to 2y.

$$6y+15=45$$

Subtract 15 from both sides of the equation.

$$6y=30$$

Divide both sides by 6.

$$y=5$$

Substitute 5 for y in x=\frac{4}{5}y+3. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{5}\times 5+3$$

Multiply \frac{4}{5} times 5.

$$x=4+3$$

Add 3 to 4.

$$x=7$$

The system is now solved.

$$x=7,y=5$$

Steps Using Matrices

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

$$3\left(3x-4y+3\right)=48-4\left(4x-2y-9\right)$$

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

$$9x-12y+9=48-4\left(4x-2y-9\right)$$

Use the distributive property to multiply -4 by 4x-2y-9.

$$9x-12y+9=48-16x+8y+36$$

Add 48 and 36 to get 84.

$$9x-12y+9=84-16x+8y$$

Add 16x to both sides.

$$9x-12y+9+16x=84+8y$$

Combine 9x and 16x to get 25x.

$$25x-12y+9=84+8y$$

Subtract 8y from both sides.

$$25x-12y+9-8y=84$$

Combine -12y and -8y to get -20y.

$$25x-20y+9=84$$

Subtract 9 from both sides.

$$25x-20y=84-9$$

Subtract 9 from 84 to get 75.

$$25x-20y=75$$

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

$$4\left(2x-y+3\right)=48+3\left(x-2y+3\right)$$

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

$$8x-4y+12=48+3\left(x-2y+3\right)$$

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

$$8x-4y+12=48+3x-6y+9$$

Add 48 and 9 to get 57.

$$8x-4y+12=57+3x-6y$$

Subtract 3x from both sides.

$$8x-4y+12-3x=57-6y$$

Combine 8x and -3x to get 5x.

$$5x-4y+12=57-6y$$

Add 6y to both sides.

$$5x-4y+12+6y=57$$

Combine -4y and 6y to get 2y.

$$5x+2y+12=57$$

Subtract 12 from both sides.

$$5x+2y=57-12$$

Subtract 12 from 57 to get 45.

$$5x+2y=45$$

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

$$25x-20y=75,5x+2y=45$$

Write the equations in matrix form.

$$\left(\begin{matrix}25&-20\\5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}75\\45\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}25&-20\\5&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}25&-20\\5&2\end{matrix}\right))\left(\begin{matrix}25&-20\\5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}25&-20\\5&2\end{matrix}\right))\left(\begin{matrix}75\\45\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}25&-20\\5&2\end{matrix}\right))\left(\begin{matrix}75\\45\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}25&-20\\5&2\end{matrix}\right))\left(\begin{matrix}75\\45\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{25\times 2-\left(-20\times 5\right)}&-\frac{-20}{25\times 2-\left(-20\times 5\right)}\\-\frac{5}{25\times 2-\left(-20\times 5\right)}&\frac{25}{25\times 2-\left(-20\times 5\right)}\end{matrix}\right)\left(\begin{matrix}75\\45\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{75}&\frac{2}{15}\\-\frac{1}{30}&\frac{1}{6}\end{matrix}\right)\left(\begin{matrix}75\\45\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{75}\times 75+\frac{2}{15}\times 45\\-\frac{1}{30}\times 75+\frac{1}{6}\times 45\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=7,y=5$$

Steps Using Elimination

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

$$3\left(3x-4y+3\right)=48-4\left(4x-2y-9\right)$$

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

$$9x-12y+9=48-4\left(4x-2y-9\right)$$

Use the distributive property to multiply -4 by 4x-2y-9.

$$9x-12y+9=48-16x+8y+36$$

Add 48 and 36 to get 84.

$$9x-12y+9=84-16x+8y$$

Add 16x to both sides.

$$9x-12y+9+16x=84+8y$$

Combine 9x and 16x to get 25x.

$$25x-12y+9=84+8y$$

Subtract 8y from both sides.

$$25x-12y+9-8y=84$$

Combine -12y and -8y to get -20y.

$$25x-20y+9=84$$

Subtract 9 from both sides.

$$25x-20y=84-9$$

Subtract 9 from 84 to get 75.

$$25x-20y=75$$

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

$$4\left(2x-y+3\right)=48+3\left(x-2y+3\right)$$

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

$$8x-4y+12=48+3\left(x-2y+3\right)$$

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

$$8x-4y+12=48+3x-6y+9$$

Add 48 and 9 to get 57.

$$8x-4y+12=57+3x-6y$$

Subtract 3x from both sides.

$$8x-4y+12-3x=57-6y$$

Combine 8x and -3x to get 5x.

$$5x-4y+12=57-6y$$

Add 6y to both sides.

$$5x-4y+12+6y=57$$

Combine -4y and 6y to get 2y.

$$5x+2y+12=57$$

Subtract 12 from both sides.

$$5x+2y=57-12$$

Subtract 12 from 57 to get 45.

$$5x+2y=45$$

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.

$$25x-20y=75,5x+2y=45$$

To make 25x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 25.

$$5\times 25x+5\left(-20\right)y=5\times 75,25\times 5x+25\times 2y=25\times 45$$

Simplify.

$$125x-100y=375,125x+50y=1125$$

Subtract 125x+50y=1125 from 125x-100y=375 by subtracting like terms on each side of the equal sign.

$$125x-125x-100y-50y=375-1125$$

Add 125x to -125x. Terms 125x and -125x cancel out, leaving an equation with only one variable that can be solved.

$$-100y-50y=375-1125$$

Add -100y to -50y.

$$-150y=375-1125$$

Add 375 to -1125.

$$-150y=-750$$

Divide both sides by -150.

$$y=5$$

Substitute 5 for y in 5x+2y=45. Because the resulting equation contains only one variable, you can solve for x directly.

$$5x+2\times 5=45$$

Multiply 2 times 5.

$$5x+10=45$$

Subtract 10 from both sides of the equation.

$$5x=35$$

Divide both sides by 5.

$$x=7$$

The system is now solved.

$$x=7,y=5$$