Solve {l}{3(5x-2)-7(2y+3)=2}{2(3x-y)-23=3(4-9x)}

$$\left\{ \begin{array} { l } { 3 ( 5 x - 2 ) - 7 ( 2 y + 3 ) = 2 } \\ { 2 ( 3 x - y ) - 23 = 3 ( 4 - 9 x ) } \end{array} \right.$$

Solve for x, y

$$x=1$$

$$y=-1$$

Steps Using Substitution

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

$$15x-6-7\left(2y+3\right)=2$$

Use the distributive property to multiply -7 by 2y+3.

$$15x-6-14y-21=2$$

Subtract 21 from -6 to get -27.

$$15x-27-14y=2$$

Add 27 to both sides.

$$15x-14y=2+27$$

Add 2 and 27 to get 29.

$$15x-14y=29$$

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

$$6x-2y-23=3\left(4-9x\right)$$

Use the distributive property to multiply 3 by 4-9x.

$$6x-2y-23=12-27x$$

Add 27x to both sides.

$$6x-2y-23+27x=12$$

Combine 6x and 27x to get 33x.

$$33x-2y-23=12$$

Add 23 to both sides.

$$33x-2y=12+23$$

Add 12 and 23 to get 35.

$$33x-2y=35$$

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.

$$15x-14y=29,33x-2y=35$$

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

$$15x-14y=29$$

Add 14y to both sides of the equation.

$$15x=14y+29$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(14y+29\right)$$

Multiply \frac{1}{15} times 14y+29.

$$x=\frac{14}{15}y+\frac{29}{15}$$

Substitute \frac{14y+29}{15} for x in the other equation, 33x-2y=35.

$$33\left(\frac{14}{15}y+\frac{29}{15}\right)-2y=35$$

Multiply 33 times \frac{14y+29}{15}.

$$\frac{154}{5}y+\frac{319}{5}-2y=35$$

Add \frac{154y}{5} to -2y.

$$\frac{144}{5}y+\frac{319}{5}=35$$

Subtract \frac{319}{5} from both sides of the equation.

$$\frac{144}{5}y=-\frac{144}{5}$$

Divide both sides of the equation by \frac{144}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-1$$

Substitute -1 for y in x=\frac{14}{15}y+\frac{29}{15}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{14}{15}\left(-1\right)+\frac{29}{15}$$

Multiply \frac{14}{15} times -1.

$$x=\frac{-14+29}{15}$$

Add \frac{29}{15} to -\frac{14}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=1$$

The system is now solved.

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

Steps Using Matrices

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

$$15x-6-7\left(2y+3\right)=2$$

Use the distributive property to multiply -7 by 2y+3.

$$15x-6-14y-21=2$$

Subtract 21 from -6 to get -27.

$$15x-27-14y=2$$

Add 27 to both sides.

$$15x-14y=2+27$$

Add 2 and 27 to get 29.

$$15x-14y=29$$

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

$$6x-2y-23=3\left(4-9x\right)$$

Use the distributive property to multiply 3 by 4-9x.

$$6x-2y-23=12-27x$$

Add 27x to both sides.

$$6x-2y-23+27x=12$$

Combine 6x and 27x to get 33x.

$$33x-2y-23=12$$

Add 23 to both sides.

$$33x-2y=12+23$$

Add 12 and 23 to get 35.

$$33x-2y=35$$

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

$$15x-14y=29,33x-2y=35$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&-14\\33&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}29\\35\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&-14\\33&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&-14\\33&-2\end{matrix}\right))\left(\begin{matrix}15&-14\\33&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-14\\33&-2\end{matrix}\right))\left(\begin{matrix}29\\35\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-14\\33&-2\end{matrix}\right))\left(\begin{matrix}29\\35\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&-14\\33&-2\end{matrix}\right))\left(\begin{matrix}29\\35\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{15\left(-2\right)-\left(-14\times 33\right)}&-\frac{-14}{15\left(-2\right)-\left(-14\times 33\right)}\\-\frac{33}{15\left(-2\right)-\left(-14\times 33\right)}&\frac{15}{15\left(-2\right)-\left(-14\times 33\right)}\end{matrix}\right)\left(\begin{matrix}29\\35\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{216}&\frac{7}{216}\\-\frac{11}{144}&\frac{5}{144}\end{matrix}\right)\left(\begin{matrix}29\\35\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{216}\times 29+\frac{7}{216}\times 35\\-\frac{11}{144}\times 29+\frac{5}{144}\times 35\end{matrix}\right)$$

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 3 by 5x-2.

$$15x-6-7\left(2y+3\right)=2$$

Use the distributive property to multiply -7 by 2y+3.

$$15x-6-14y-21=2$$

Subtract 21 from -6 to get -27.

$$15x-27-14y=2$$

Add 27 to both sides.

$$15x-14y=2+27$$

Add 2 and 27 to get 29.

$$15x-14y=29$$

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

$$6x-2y-23=3\left(4-9x\right)$$

Use the distributive property to multiply 3 by 4-9x.

$$6x-2y-23=12-27x$$

Add 27x to both sides.

$$6x-2y-23+27x=12$$

Combine 6x and 27x to get 33x.

$$33x-2y-23=12$$

Add 23 to both sides.

$$33x-2y=12+23$$

Add 12 and 23 to get 35.

$$33x-2y=35$$

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.

$$15x-14y=29,33x-2y=35$$

To make 15x and 33x equal, multiply all terms on each side of the first equation by 33 and all terms on each side of the second by 15.

$$33\times 15x+33\left(-14\right)y=33\times 29,15\times 33x+15\left(-2\right)y=15\times 35$$

Simplify.

$$495x-462y=957,495x-30y=525$$

Subtract 495x-30y=525 from 495x-462y=957 by subtracting like terms on each side of the equal sign.

$$495x-495x-462y+30y=957-525$$

Add 495x to -495x. Terms 495x and -495x cancel out, leaving an equation with only one variable that can be solved.

$$-462y+30y=957-525$$

Add -462y to 30y.

$$-432y=957-525$$

Add 957 to -525.

$$-432y=432$$

Divide both sides by -432.

$$y=-1$$

Substitute -1 for y in 33x-2y=35. Because the resulting equation contains only one variable, you can solve for x directly.

$$33x-2\left(-1\right)=35$$

Multiply -2 times -1.

$$33x+2=35$$

Subtract 2 from both sides of the equation.

$$33x=33$$

Divide both sides by 33.

$$x=1$$

The system is now solved.

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