Solve {l}{2x+4y=14}{10x+5y=40}

$$\left. \begin{array} { l } { 2 x + 4 y = 14 } \\ { 10 x + 5 y = 40 } \end{array} \right.$$

Solve for x, y

$$x=3$$

$$y=2$$

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.

$$2x+4y=14,10x+5y=40$$

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

$$2x+4y=14$$

Subtract 4y from both sides of the equation.

$$2x=-4y+14$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-4y+14\right)$$

Multiply \frac{1}{2} times -4y+14.

$$x=-2y+7$$

Substitute -2y+7 for x in the other equation, 10x+5y=40.

$$10\left(-2y+7\right)+5y=40$$

Multiply 10 times -2y+7.

$$-20y+70+5y=40$$

Add -20y to 5y.

$$-15y+70=40$$

Subtract 70 from both sides of the equation.

$$-15y=-30$$

Divide both sides by -15.

$$y=2$$

Substitute 2 for y in x=-2y+7. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-2\times 2+7$$

Multiply -2 times 2.

$$x=-4+7$$

Add 7 to -4.

$$x=3$$

The system is now solved.

$$x=3,y=2$$

Steps Using Matrices

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

$$2x+4y=14,10x+5y=40$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&4\\10&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&4\\10&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&4\\10&5\end{matrix}\right))\left(\begin{matrix}2&4\\10&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\10&5\end{matrix}\right))\left(\begin{matrix}14\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\10&5\end{matrix}\right))\left(\begin{matrix}14\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\10&5\end{matrix}\right))\left(\begin{matrix}14\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{2\times 5-4\times 10}&-\frac{4}{2\times 5-4\times 10}\\-\frac{10}{2\times 5-4\times 10}&\frac{2}{2\times 5-4\times 10}\end{matrix}\right)\left(\begin{matrix}14\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{6}&\frac{2}{15}\\\frac{1}{3}&-\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}14\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{6}\times 14+\frac{2}{15}\times 40\\\frac{1}{3}\times 14-\frac{1}{15}\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=3,y=2$$

Steps Using Elimination

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.

$$2x+4y=14,10x+5y=40$$

To make 2x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 2.

$$10\times 2x+10\times 4y=10\times 14,2\times 10x+2\times 5y=2\times 40$$

Simplify.

$$20x+40y=140,20x+10y=80$$

Subtract 20x+10y=80 from 20x+40y=140 by subtracting like terms on each side of the equal sign.

$$20x-20x+40y-10y=140-80$$

Add 20x to -20x. Terms 20x and -20x cancel out, leaving an equation with only one variable that can be solved.

$$40y-10y=140-80$$

Add 40y to -10y.

$$30y=140-80$$

Add 140 to -80.

$$30y=60$$

Divide both sides by 30.

$$y=2$$

Substitute 2 for y in 10x+5y=40. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+5\times 2=40$$

Multiply 5 times 2.

$$10x+10=40$$

Subtract 10 from both sides of the equation.

$$10x=30$$

Divide both sides by 10.

$$x=3$$

The system is now solved.

$$x=3,y=2$$