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

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

Solve for y, x

$$x=\frac{30}{47}\approx 0.638297872$$

$$y=\frac{3}{47}\approx 0.063829787$$

Steps Using Substitution

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

$$20y+8y=3x-2y$$

Combine 20y and 8y to get 28y.

$$28y=3x-2y$$

Subtract 3x from both sides.

$$28y-3x=-2y$$

Add 2y to both sides.

$$28y-3x+2y=0$$

Combine 28y and 2y to get 30y.

$$30y-3x=0$$

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

$$y+4x=3-6y$$

Add 6y to both sides.

$$y+4x+6y=3$$

Combine y and 6y to get 7y.

$$7y+4x=3$$

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.

$$30y-3x=0,7y+4x=3$$

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

$$30y-3x=0$$

Add 3x to both sides of the equation.

$$30y=3x$$

Divide both sides by 30.

$$y=\frac{1}{30}\times 3x$$

Multiply \frac{1}{30} times 3x.

$$y=\frac{1}{10}x$$

Substitute \frac{x}{10} for y in the other equation, 7y+4x=3.

$$7\times \frac{1}{10}x+4x=3$$

Multiply 7 times \frac{x}{10}.

$$\frac{7}{10}x+4x=3$$

Add \frac{7x}{10} to 4x.

$$\frac{47}{10}x=3$$

Divide both sides of the equation by \frac{47}{10}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=\frac{30}{47}$$

Substitute \frac{30}{47} for x in y=\frac{1}{10}x. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{1}{10}\times \frac{30}{47}$$

Multiply \frac{1}{10} times \frac{30}{47} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$y=\frac{3}{47}$$

The system is now solved.

$$y=\frac{3}{47},x=\frac{30}{47}$$

Steps Using Matrices

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

$$20y+8y=3x-2y$$

Combine 20y and 8y to get 28y.

$$28y=3x-2y$$

Subtract 3x from both sides.

$$28y-3x=-2y$$

Add 2y to both sides.

$$28y-3x+2y=0$$

Combine 28y and 2y to get 30y.

$$30y-3x=0$$

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

$$y+4x=3-6y$$

Add 6y to both sides.

$$y+4x+6y=3$$

Combine y and 6y to get 7y.

$$7y+4x=3$$

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

$$30y-3x=0,7y+4x=3$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&-3\\7&4\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&-3\\7&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&-3\\7&4\end{matrix}\right))\left(\begin{matrix}30&-3\\7&4\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}30&-3\\7&4\end{matrix}\right))\left(\begin{matrix}0\\3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}30&-3\\7&4\end{matrix}\right))\left(\begin{matrix}0\\3\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}30&-3\\7&4\end{matrix}\right))\left(\begin{matrix}0\\3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{4}{30\times 4-\left(-3\times 7\right)}&-\frac{-3}{30\times 4-\left(-3\times 7\right)}\\-\frac{7}{30\times 4-\left(-3\times 7\right)}&\frac{30}{30\times 4-\left(-3\times 7\right)}\end{matrix}\right)\left(\begin{matrix}0\\3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{4}{141}&\frac{1}{47}\\-\frac{7}{141}&\frac{10}{47}\end{matrix}\right)\left(\begin{matrix}0\\3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{47}\times 3\\\frac{10}{47}\times 3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{3}{47}\\\frac{30}{47}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=\frac{3}{47},x=\frac{30}{47}$$

Steps Using Elimination

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

$$20y+8y=3x-2y$$

Combine 20y and 8y to get 28y.

$$28y=3x-2y$$

Subtract 3x from both sides.

$$28y-3x=-2y$$

Add 2y to both sides.

$$28y-3x+2y=0$$

Combine 28y and 2y to get 30y.

$$30y-3x=0$$

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

$$y+4x=3-6y$$

Add 6y to both sides.

$$y+4x+6y=3$$

Combine y and 6y to get 7y.

$$7y+4x=3$$

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.

$$30y-3x=0,7y+4x=3$$

To make 30y and 7y equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 30.

$$7\times 30y+7\left(-3\right)x=0,30\times 7y+30\times 4x=30\times 3$$

Simplify.

$$210y-21x=0,210y+120x=90$$

Subtract 210y+120x=90 from 210y-21x=0 by subtracting like terms on each side of the equal sign.

$$210y-210y-21x-120x=-90$$

Add 210y to -210y. Terms 210y and -210y cancel out, leaving an equation with only one variable that can be solved.

$$-21x-120x=-90$$

Add -21x to -120x.

$$-141x=-90$$

Divide both sides by -141.

$$x=\frac{30}{47}$$

Substitute \frac{30}{47} for x in 7y+4x=3. Because the resulting equation contains only one variable, you can solve for y directly.

$$7y+4\times \frac{30}{47}=3$$

Multiply 4 times \frac{30}{47}.

$$7y+\frac{120}{47}=3$$

Subtract \frac{120}{47} from both sides of the equation.

$$7y=\frac{21}{47}$$

Divide both sides by 7.

$$y=\frac{3}{47}$$

The system is now solved.

$$y=\frac{3}{47},x=\frac{30}{47}$$