Solve {l}{5x-4y=0}{2x+4y/5=1-2x}

$$\left\{ \begin{array} { l } { 5 x - 4 y = 0 } \\ { 2 x + \frac { 4 y } { 5 } = 1 - 2 x } \end{array} \right.$$

Solve for x, y

$$x=\frac{1}{5}=0.2$$

$$y=\frac{1}{4}=0.25$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 5.

$$10x+4y=5-10x$$

Add 10x to both sides.

$$10x+4y+10x=5$$

Combine 10x and 10x to get 20x.

$$20x+4y=5$$

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.

$$5x-4y=0,20x+4y=5$$

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

$$5x-4y=0$$

Add 4y to both sides of the equation.

$$5x=4y$$

Divide both sides by 5.

$$x=\frac{1}{5}\times 4y$$

Multiply \frac{1}{5} times 4y.

$$x=\frac{4}{5}y$$

Substitute \frac{4y}{5} for x in the other equation, 20x+4y=5.

$$20\times \frac{4}{5}y+4y=5$$

Multiply 20 times \frac{4y}{5}.

$$16y+4y=5$$

Add 16y to 4y.

$$20y=5$$

Divide both sides by 20.

$$y=\frac{1}{4}$$

Substitute \frac{1}{4} for y in x=\frac{4}{5}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{5}\times \frac{1}{4}$$

Multiply \frac{4}{5} times \frac{1}{4} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{1}{5}$$

The system is now solved.

$$x=\frac{1}{5},y=\frac{1}{4}$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 5.

$$10x+4y=5-10x$$

Add 10x to both sides.

$$10x+4y+10x=5$$

Combine 10x and 10x to get 20x.

$$20x+4y=5$$

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

$$5x-4y=0,20x+4y=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-4\\20&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-4\\20&4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-4\\20&4\end{matrix}\right))\left(\begin{matrix}5&-4\\20&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-4\\20&4\end{matrix}\right))\left(\begin{matrix}0\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-4\\20&4\end{matrix}\right))\left(\begin{matrix}0\\5\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-4\\20&4\end{matrix}\right))\left(\begin{matrix}0\\5\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5\times 4-\left(-4\times 20\right)}&-\frac{-4}{5\times 4-\left(-4\times 20\right)}\\-\frac{20}{5\times 4-\left(-4\times 20\right)}&\frac{5}{5\times 4-\left(-4\times 20\right)}\end{matrix}\right)\left(\begin{matrix}0\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{25}&\frac{1}{25}\\-\frac{1}{5}&\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}0\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{25}\times 5\\\frac{1}{20}\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\\\frac{1}{4}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1}{5},y=\frac{1}{4}$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 5.

$$10x+4y=5-10x$$

Add 10x to both sides.

$$10x+4y+10x=5$$

Combine 10x and 10x to get 20x.

$$20x+4y=5$$

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.

$$5x-4y=0,20x+4y=5$$

To make 5x and 20x equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 5.

$$20\times 5x+20\left(-4\right)y=0,5\times 20x+5\times 4y=5\times 5$$

Simplify.

$$100x-80y=0,100x+20y=25$$

Subtract 100x+20y=25 from 100x-80y=0 by subtracting like terms on each side of the equal sign.

$$100x-100x-80y-20y=-25$$

Add 100x to -100x. Terms 100x and -100x cancel out, leaving an equation with only one variable that can be solved.

$$-80y-20y=-25$$

Add -80y to -20y.

$$-100y=-25$$

Divide both sides by -100.

$$y=\frac{1}{4}$$

Substitute \frac{1}{4} for y in 20x+4y=5. Because the resulting equation contains only one variable, you can solve for x directly.

$$20x+4\times \frac{1}{4}=5$$

Multiply 4 times \frac{1}{4}.

$$20x+1=5$$

Subtract 1 from both sides of the equation.

$$20x=4$$

Divide both sides by 20.

$$x=\frac{1}{5}$$

The system is now solved.

$$x=\frac{1}{5},y=\frac{1}{4}$$