Solve l_{1}:3y=4x+10;l_{2}:4y+5x=4

$$l _ { 1 } : 3 y = 4 x + 10 ; l _ { 2 } : 4 y + 5 x = 4$$

Solve for x, y

$$x=-\frac{15l_{2}-8l_{1}}{2\left(5l_{1}+3l_{2}\right)}$$

$$y=\frac{198}{5l_{1}+3l_{2}}$$

$$l_{1}\neq -\frac{3l_{2}}{5}$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 3.

$$l_{1}y=12x+30$$

Divide both sides by l_{1}.

$$y=\frac{1}{l_{1}}\left(12x+30\right)$$

Multiply \frac{1}{l_{1}} times 12x+30.

$$y=\frac{12}{l_{1}}x+\frac{30}{l_{1}}$$

Substitute \frac{6\left(5+2x\right)}{l_{1}} for y in the other equation, l_{2}y+20x=16.

$$l_{2}\left(\frac{12}{l_{1}}x+\frac{30}{l_{1}}\right)+20x=16$$

Multiply l_{2} times \frac{6\left(5+2x\right)}{l_{1}}.

$$\frac{12l_{2}}{l_{1}}x+\frac{30l_{2}}{l_{1}}+20x=16$$

Add \frac{12l_{2}x}{l_{1}} to 20x.

$$\left(\frac{12l_{2}}{l_{1}}+20\right)x+\frac{30l_{2}}{l_{1}}=16$$

Subtract \frac{30l_{2}}{l_{1}} from both sides of the equation.

$$\left(\frac{12l_{2}}{l_{1}}+20\right)x=-\frac{30l_{2}}{l_{1}}+16$$

Divide both sides by 20+\frac{12l_{2}}{l_{1}}.

$$x=\frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)}$$

Substitute \frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)} for x in y=\frac{12}{l_{1}}x+\frac{30}{l_{1}}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{12}{l_{1}}\times \frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)}+\frac{30}{l_{1}}$$

Multiply \frac{12}{l_{1}} times \frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)}.

$$y=\frac{6\left(8l_{1}-15l_{2}\right)}{l_{1}\left(5l_{1}+3l_{2}\right)}+\frac{30}{l_{1}}$$

Add \frac{30}{l_{1}} to \frac{6\left(8l_{1}-15l_{2}\right)}{l_{1}\left(5l_{1}+3l_{2}\right)}.

$$y=\frac{198}{5l_{1}+3l_{2}}$$

The system is now solved.

$$y=\frac{198}{5l_{1}+3l_{2}},x=\frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 3.

$$l_{1}y=12x+30$$

Subtract 12x from both sides.

$$l_{1}y-12x=30$$

Consider the second equation. Multiply both sides of the equation by 4.

$$l_{2}y+20x=16$$

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

$$l_{1}y-12x=30,l_{2}y+20x=16$$

Write the equations in matrix form.

$$\left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}30\\16\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right).

$$inverse(\left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right))\left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right))\left(\begin{matrix}30\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right))\left(\begin{matrix}30\\16\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}l_{1}&-12\\l_{2}&20\end{matrix}\right))\left(\begin{matrix}30\\16\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{20}{l_{1}\times 20-\left(-12l_{2}\right)}&-\frac{-12}{l_{1}\times 20-\left(-12l_{2}\right)}\\-\frac{l_{2}}{l_{1}\times 20-\left(-12l_{2}\right)}&\frac{l_{1}}{l_{1}\times 20-\left(-12l_{2}\right)}\end{matrix}\right)\left(\begin{matrix}30\\16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5l_{1}+3l_{2}}&\frac{3}{5l_{1}+3l_{2}}\\-\frac{l_{2}}{4\left(5l_{1}+3l_{2}\right)}&\frac{l_{1}}{4\left(5l_{1}+3l_{2}\right)}\end{matrix}\right)\left(\begin{matrix}30\\16\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5l_{1}+3l_{2}}\times 30+\frac{3}{5l_{1}+3l_{2}}\times 16\\\left(-\frac{l_{2}}{4\left(5l_{1}+3l_{2}\right)}\right)\times 30+\frac{l_{1}}{4\left(5l_{1}+3l_{2}\right)}\times 16\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{198}{5l_{1}+3l_{2}}\\\frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{198}{5l_{1}+3l_{2}},x=\frac{8l_{1}-15l_{2}}{2\left(5l_{1}+3l_{2}\right)}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 3.

$$l_{1}y=12x+30$$

Subtract 12x from both sides.

$$l_{1}y-12x=30$$

Consider the second equation. Multiply both sides of the equation by 4.

$$l_{2}y+20x=16$$

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.

$$l_{1}y-12x=30,l_{2}y+20x=16$$

To make l_{1}y and l_{2}y equal, multiply all terms on each side of the first equation by l_{2} and all terms on each side of the second by l_{1}.

$$l_{2}l_{1}y+l_{2}\left(-12\right)x=l_{2}\times 30,l_{1}l_{2}y+l_{1}\times 20x=l_{1}\times 16$$

Simplify.

$$l_{1}l_{2}y+\left(-12l_{2}\right)x=30l_{2},l_{1}l_{2}y+20l_{1}x=16l_{1}$$

Subtract l_{1}l_{2}y+20l_{1}x=16l_{1} from l_{1}l_{2}y+\left(-12l_{2}\right)x=30l_{2} by subtracting like terms on each side of the equal sign.

$$l_{1}l_{2}y+\left(-l_{1}l_{2}\right)y+\left(-12l_{2}\right)x+\left(-20l_{1}\right)x=30l_{2}-16l_{1}$$

Add l_{2}l_{1}y to -l_{2}l_{1}y. Terms l_{2}l_{1}y and -l_{2}l_{1}y cancel out, leaving an equation with only one variable that can be solved.

$$\left(-12l_{2}\right)x+\left(-20l_{1}\right)x=30l_{2}-16l_{1}$$

Add -12l_{2}x to -20l_{1}x.

$$\left(-20l_{1}-12l_{2}\right)x=30l_{2}-16l_{1}$$

Divide both sides by -12l_{2}-20l_{1}.

$$x=-\frac{15l_{2}-8l_{1}}{2\left(5l_{1}+3l_{2}\right)}$$

Substitute -\frac{15l_{2}-8l_{1}}{2\left(3l_{2}+5l_{1}\right)} for x in l_{2}y+20x=16. Because the resulting equation contains only one variable, you can solve for y directly.

$$l_{2}y+20\left(-\frac{15l_{2}-8l_{1}}{2\left(5l_{1}+3l_{2}\right)}\right)=16$$

Multiply 20 times -\frac{15l_{2}-8l_{1}}{2\left(3l_{2}+5l_{1}\right)}.

$$l_{2}y-\frac{10\left(15l_{2}-8l_{1}\right)}{5l_{1}+3l_{2}}=16$$

Add \frac{10\left(15l_{2}-8l_{1}\right)}{3l_{2}+5l_{1}} to both sides of the equation.

$$l_{2}y=\frac{198l_{2}}{5l_{1}+3l_{2}}$$

Divide both sides by l_{2}.

$$y=\frac{198}{5l_{1}+3l_{2}}$$

The system is now solved.

$$y=\frac{198}{5l_{1}+3l_{2}},x=-\frac{15l_{2}-8l_{1}}{2\left(5l_{1}+3l_{2}\right)}$$