Solve {l}{3(y-2/3)-4(x+5)=14}{7(y-2/3)+2(x+5)=10}

$$\left\{ \begin{array} { l } { 3 ( y - \frac { 2 } { 3 } ) - 4 ( x + 5 ) = 14 } \\ { 7 ( y - \frac { 2 } { 3 } ) + 2 ( x + 5 ) = 10 } \end{array} \right.$$

Solve for y, x

$$x=-7$$

$$y = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using Substitution

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.

$$3\left(y-\frac{2}{3}\right)-4\left(x+5\right)=14,7\left(y-\frac{2}{3}\right)+2\left(x+5\right)=10$$

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

$$3\left(y-\frac{2}{3}\right)-4\left(x+5\right)=14$$

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

$$3y-2-4\left(x+5\right)=14$$

Multiply -4 times x+5.

$$3y-2-4x-20=14$$

Add -2 to -20.

$$3y-4x-22=14$$

Add 22 to both sides of the equation.

$$3y-4x=36$$

Add 4x to both sides of the equation.

$$3y=4x+36$$

Divide both sides by 3.

$$y=\frac{1}{3}\left(4x+36\right)$$

Multiply \frac{1}{3} times 36+4x.

$$y=\frac{4}{3}x+12$$

Substitute \frac{4x}{3}+12 for y in the other equation, 7\left(y-\frac{2}{3}\right)+2\left(x+5\right)=10.

$$7\left(\frac{4}{3}x+12-\frac{2}{3}\right)+2\left(x+5\right)=10$$

Add 12 to -\frac{2}{3}.

$$7\left(\frac{4}{3}x+\frac{34}{3}\right)+2\left(x+5\right)=10$$

Multiply 7 times \frac{4x+34}{3}.

$$\frac{28}{3}x+\frac{238}{3}+2\left(x+5\right)=10$$

Multiply 2 times x+5.

$$\frac{28}{3}x+\frac{238}{3}+2x+10=10$$

Add \frac{28x}{3} to 2x.

$$\frac{34}{3}x+\frac{238}{3}+10=10$$

Add \frac{238}{3} to 10.

$$\frac{34}{3}x+\frac{268}{3}=10$$

Subtract \frac{268}{3} from both sides of the equation.

$$\frac{34}{3}x=-\frac{238}{3}$$

Divide both sides of the equation by \frac{34}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=-7$$

Substitute -7 for x in y=\frac{4}{3}x+12. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{4}{3}\left(-7\right)+12$$

Multiply \frac{4}{3} times -7.

$$y=-\frac{28}{3}+12$$

Add 12 to -\frac{28}{3}.

$$y=\frac{8}{3}$$

The system is now solved.

$$y=\frac{8}{3},x=-7$$

Steps Using Matrices

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

$$3\left(y-\frac{2}{3}\right)-4\left(x+5\right)=14,7\left(y-\frac{2}{3}\right)+2\left(x+5\right)=10$$

Simplify the first equation to put it in standard form.

$$3\left(y-\frac{2}{3}\right)-4\left(x+5\right)=14$$

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

$$3y-2-4\left(x+5\right)=14$$

Multiply -4 times x+5.

$$3y-2-4x-20=14$$

Add -2 to -20.

$$3y-4x-22=14$$

Add 22 to both sides of the equation.

$$3y-4x=36$$

Simplify the second equation to put it in standard form.

$$7\left(y-\frac{2}{3}\right)+2\left(x+5\right)=10$$

Multiply 7 times y-\frac{2}{3}.

$$7y-\frac{14}{3}+2\left(x+5\right)=10$$

Multiply 2 times x+5.

$$7y-\frac{14}{3}+2x+10=10$$

Add -\frac{14}{3} to 10.

$$7y+2x+\frac{16}{3}=10$$

Subtract \frac{16}{3} from both sides of the equation.

$$7y+2x=\frac{14}{3}$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-4\\7&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}36\\\frac{14}{3}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-4\\7&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-4\\7&2\end{matrix}\right))\left(\begin{matrix}3&-4\\7&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\7&2\end{matrix}\right))\left(\begin{matrix}36\\\frac{14}{3}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\7&2\end{matrix}\right))\left(\begin{matrix}36\\\frac{14}{3}\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\7&2\end{matrix}\right))\left(\begin{matrix}36\\\frac{14}{3}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3\times 2-\left(-4\times 7\right)}&-\frac{-4}{3\times 2-\left(-4\times 7\right)}\\-\frac{7}{3\times 2-\left(-4\times 7\right)}&\frac{3}{3\times 2-\left(-4\times 7\right)}\end{matrix}\right)\left(\begin{matrix}36\\\frac{14}{3}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{17}&\frac{2}{17}\\-\frac{7}{34}&\frac{3}{34}\end{matrix}\right)\left(\begin{matrix}36\\\frac{14}{3}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{17}\times 36+\frac{2}{17}\times \frac{14}{3}\\-\frac{7}{34}\times 36+\frac{3}{34}\times \frac{14}{3}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{8}{3}\\-7\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{8}{3},x=-7$$