Solve {l}{3(2x+1)-5(y-3)=1}{5(1-x)-4(2y+1)=3}

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

Solve for x, y

$$x=-2$$

$$y=1$$

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.

$$3\left(2x+1\right)-5\left(y-3\right)=1,5\left(-x+1\right)-4\left(2y+1\right)=3$$

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

$$3\left(2x+1\right)-5\left(y-3\right)=1$$

Multiply 3 times 2x+1.

$$6x+3-5\left(y-3\right)=1$$

Multiply -5 times y-3.

$$6x+3-5y+15=1$$

Add 3 to 15.

$$6x-5y+18=1$$

Subtract 18 from both sides of the equation.

$$6x-5y=-17$$

Add 5y to both sides of the equation.

$$6x=5y-17$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(5y-17\right)$$

Multiply \frac{1}{6} times 5y-17.

$$x=\frac{5}{6}y-\frac{17}{6}$$

Substitute \frac{5y-17}{6} for x in the other equation, 5\left(-x+1\right)-4\left(2y+1\right)=3.

$$5\left(-\left(\frac{5}{6}y-\frac{17}{6}\right)+1\right)-4\left(2y+1\right)=3$$

Multiply -1 times \frac{5y-17}{6}.

$$5\left(-\frac{5}{6}y+\frac{17}{6}+1\right)-4\left(2y+1\right)=3$$

Add \frac{17}{6} to 1.

$$5\left(-\frac{5}{6}y+\frac{23}{6}\right)-4\left(2y+1\right)=3$$

Multiply 5 times \frac{-5y+23}{6}.

$$-\frac{25}{6}y+\frac{115}{6}-4\left(2y+1\right)=3$$

Multiply -4 times 2y+1.

$$-\frac{25}{6}y+\frac{115}{6}-8y-4=3$$

Add -\frac{25y}{6} to -8y.

$$-\frac{73}{6}y+\frac{115}{6}-4=3$$

Add \frac{115}{6} to -4.

$$-\frac{73}{6}y+\frac{91}{6}=3$$

Subtract \frac{91}{6} from both sides of the equation.

$$-\frac{73}{6}y=-\frac{73}{6}$$

Divide both sides of the equation by -\frac{73}{6}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=1$$

Substitute 1 for y in x=\frac{5}{6}y-\frac{17}{6}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5-17}{6}$$

Add -\frac{17}{6} to \frac{5}{6} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-2$$

The system is now solved.

$$x=-2,y=1$$

Steps Using Matrices

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

$$3\left(2x+1\right)-5\left(y-3\right)=1,5\left(-x+1\right)-4\left(2y+1\right)=3$$

Simplify the first equation to put it in standard form.

$$3\left(2x+1\right)-5\left(y-3\right)=1$$

Multiply 3 times 2x+1.

$$6x+3-5\left(y-3\right)=1$$

Multiply -5 times y-3.

$$6x+3-5y+15=1$$

Add 3 to 15.

$$6x-5y+18=1$$

Subtract 18 from both sides of the equation.

$$6x-5y=-17$$

Simplify the second equation to put it in standard form.

$$5\left(-x+1\right)-4\left(2y+1\right)=3$$

Multiply 5 times -x+1.

$$-5x+5-4\left(2y+1\right)=3$$

Multiply -4 times 2y+1.

$$-5x+5-8y-4=3$$

Add 5 to -4.

$$-5x-8y+1=3$$

Subtract 1 from both sides of the equation.

$$-5x-8y=2$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-17\\2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right))\left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right))\left(\begin{matrix}-17\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right))\left(\begin{matrix}-17\\2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&-5\\-5&-8\end{matrix}\right))\left(\begin{matrix}-17\\2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{6\left(-8\right)-\left(-5\left(-5\right)\right)}&-\frac{-5}{6\left(-8\right)-\left(-5\left(-5\right)\right)}\\-\frac{-5}{6\left(-8\right)-\left(-5\left(-5\right)\right)}&\frac{6}{6\left(-8\right)-\left(-5\left(-5\right)\right)}\end{matrix}\right)\left(\begin{matrix}-17\\2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{73}&-\frac{5}{73}\\-\frac{5}{73}&-\frac{6}{73}\end{matrix}\right)\left(\begin{matrix}-17\\2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{73}\left(-17\right)-\frac{5}{73}\times 2\\-\frac{5}{73}\left(-17\right)-\frac{6}{73}\times 2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-2,y=1$$