Solve {l}{3x-7/4-2y+1/6=0}{x+2/5-5y+4/3=-2}

$$\left\{ \begin{array} { l } { \frac { 3 x - 7 } { 4 } - \frac { 2 y + 1 } { 6 } = 0 } \\ { \frac { x + 2 } { 5 } - \frac { 5 y + 4 } { 3 } = - 2 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=1$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,6.

$$3\left(3x-7\right)-2\left(2y+1\right)=0$$

Use the distributive property to multiply 3 by 3x-7.

$$9x-21-2\left(2y+1\right)=0$$

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

$$9x-21-4y-2=0$$

Subtract 2 from -21 to get -23.

$$9x-23-4y=0$$

Add 23 to both sides. Anything plus zero gives itself.

$$9x-4y=23$$

Consider the second equation. Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$3\left(x+2\right)-5\left(5y+4\right)=-30$$

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

$$3x+6-5\left(5y+4\right)=-30$$

Use the distributive property to multiply -5 by 5y+4.

$$3x+6-25y-20=-30$$

Subtract 20 from 6 to get -14.

$$3x-14-25y=-30$$

Add 14 to both sides.

$$3x-25y=-30+14$$

Add -30 and 14 to get -16.

$$3x-25y=-16$$

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.

$$9x-4y=23,3x-25y=-16$$

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

$$9x-4y=23$$

Add 4y to both sides of the equation.

$$9x=4y+23$$

Divide both sides by 9.

$$x=\frac{1}{9}\left(4y+23\right)$$

Multiply \frac{1}{9} times 4y+23.

$$x=\frac{4}{9}y+\frac{23}{9}$$

Substitute \frac{4y+23}{9} for x in the other equation, 3x-25y=-16.

$$3\left(\frac{4}{9}y+\frac{23}{9}\right)-25y=-16$$

Multiply 3 times \frac{4y+23}{9}.

$$\frac{4}{3}y+\frac{23}{3}-25y=-16$$

Add \frac{4y}{3} to -25y.

$$-\frac{71}{3}y+\frac{23}{3}=-16$$

Subtract \frac{23}{3} from both sides of the equation.

$$-\frac{71}{3}y=-\frac{71}{3}$$

Divide both sides of the equation by -\frac{71}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=1$$

Substitute 1 for y in x=\frac{4}{9}y+\frac{23}{9}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4+23}{9}$$

Add \frac{23}{9} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=3$$

The system is now solved.

$$x=3,y=1$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,6.

$$3\left(3x-7\right)-2\left(2y+1\right)=0$$

Use the distributive property to multiply 3 by 3x-7.

$$9x-21-2\left(2y+1\right)=0$$

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

$$9x-21-4y-2=0$$

Subtract 2 from -21 to get -23.

$$9x-23-4y=0$$

Add 23 to both sides. Anything plus zero gives itself.

$$9x-4y=23$$

Consider the second equation. Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$3\left(x+2\right)-5\left(5y+4\right)=-30$$

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

$$3x+6-5\left(5y+4\right)=-30$$

Use the distributive property to multiply -5 by 5y+4.

$$3x+6-25y-20=-30$$

Subtract 20 from 6 to get -14.

$$3x-14-25y=-30$$

Add 14 to both sides.

$$3x-25y=-30+14$$

Add -30 and 14 to get -16.

$$3x-25y=-16$$

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

$$9x-4y=23,3x-25y=-16$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&-4\\3&-25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}23\\-16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&-4\\3&-25\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&-4\\3&-25\end{matrix}\right))\left(\begin{matrix}9&-4\\3&-25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&-4\\3&-25\end{matrix}\right))\left(\begin{matrix}23\\-16\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&-4\\3&-25\end{matrix}\right))\left(\begin{matrix}23\\-16\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}9&-4\\3&-25\end{matrix}\right))\left(\begin{matrix}23\\-16\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{25}{9\left(-25\right)-\left(-4\times 3\right)}&-\frac{-4}{9\left(-25\right)-\left(-4\times 3\right)}\\-\frac{3}{9\left(-25\right)-\left(-4\times 3\right)}&\frac{9}{9\left(-25\right)-\left(-4\times 3\right)}\end{matrix}\right)\left(\begin{matrix}23\\-16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{213}&-\frac{4}{213}\\\frac{1}{71}&-\frac{3}{71}\end{matrix}\right)\left(\begin{matrix}23\\-16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{213}\times 23-\frac{4}{213}\left(-16\right)\\\frac{1}{71}\times 23-\frac{3}{71}\left(-16\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=1$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 12, the least common multiple of 4,6.

$$3\left(3x-7\right)-2\left(2y+1\right)=0$$

Use the distributive property to multiply 3 by 3x-7.

$$9x-21-2\left(2y+1\right)=0$$

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

$$9x-21-4y-2=0$$

Subtract 2 from -21 to get -23.

$$9x-23-4y=0$$

Add 23 to both sides. Anything plus zero gives itself.

$$9x-4y=23$$

Consider the second equation. Multiply both sides of the equation by 15, the least common multiple of 5,3.

$$3\left(x+2\right)-5\left(5y+4\right)=-30$$

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

$$3x+6-5\left(5y+4\right)=-30$$

Use the distributive property to multiply -5 by 5y+4.

$$3x+6-25y-20=-30$$

Subtract 20 from 6 to get -14.

$$3x-14-25y=-30$$

Add 14 to both sides.

$$3x-25y=-30+14$$

Add -30 and 14 to get -16.

$$3x-25y=-16$$

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.

$$9x-4y=23,3x-25y=-16$$

To make 9x 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 9.

$$3\times 9x+3\left(-4\right)y=3\times 23,9\times 3x+9\left(-25\right)y=9\left(-16\right)$$

Simplify.

$$27x-12y=69,27x-225y=-144$$

Subtract 27x-225y=-144 from 27x-12y=69 by subtracting like terms on each side of the equal sign.

$$27x-27x-12y+225y=69+144$$

Add 27x to -27x. Terms 27x and -27x cancel out, leaving an equation with only one variable that can be solved.

$$-12y+225y=69+144$$

Add -12y to 225y.

$$213y=69+144$$

Add 69 to 144.

$$213y=213$$

Divide both sides by 213.

$$y=1$$

Substitute 1 for y in 3x-25y=-16. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x-25=-16$$

Add 25 to both sides of the equation.

$$3x=9$$

Divide both sides by 3.

$$x=3$$

The system is now solved.

$$x=3,y=1$$