Solve {l}{4x+y=10}{3x+2y=15}

$$\left. \begin{array} { l } { 4 x + y = 10 } \\ { 3 x + 2 y = 15 } \end{array} \right.$$

Solve for x, y

$$x=1$$

$$y=6$$

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.

$$4x+y=10,3x+2y=15$$

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

$$4x+y=10$$

Subtract y from both sides of the equation.

$$4x=-y+10$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(-y+10\right)$$

Multiply \frac{1}{4} times -y+10.

$$x=-\frac{1}{4}y+\frac{5}{2}$$

Substitute -\frac{y}{4}+\frac{5}{2} for x in the other equation, 3x+2y=15.

$$3\left(-\frac{1}{4}y+\frac{5}{2}\right)+2y=15$$

Multiply 3 times -\frac{y}{4}+\frac{5}{2}.

$$-\frac{3}{4}y+\frac{15}{2}+2y=15$$

Add -\frac{3y}{4} to 2y.

$$\frac{5}{4}y+\frac{15}{2}=15$$

Subtract \frac{15}{2} from both sides of the equation.

$$\frac{5}{4}y=\frac{15}{2}$$

Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=6$$

Substitute 6 for y in x=-\frac{1}{4}y+\frac{5}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{4}\times 6+\frac{5}{2}$$

Multiply -\frac{1}{4} times 6.

$$x=\frac{-3+5}{2}$$

Add \frac{5}{2} to -\frac{3}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=1$$

The system is now solved.

$$x=1,y=6$$

Steps Using Matrices

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

$$4x+y=10,3x+2y=15$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&1\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\15\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&1\\3&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&1\\3&2\end{matrix}\right))\left(\begin{matrix}4&1\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\3&2\end{matrix}\right))\left(\begin{matrix}10\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\3&2\end{matrix}\right))\left(\begin{matrix}10\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&1\\3&2\end{matrix}\right))\left(\begin{matrix}10\\15\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{4\times 2-3}&-\frac{1}{4\times 2-3}\\-\frac{3}{4\times 2-3}&\frac{4}{4\times 2-3}\end{matrix}\right)\left(\begin{matrix}10\\15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}&-\frac{1}{5}\\-\frac{3}{5}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}10\\15\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{5}\times 10-\frac{1}{5}\times 15\\-\frac{3}{5}\times 10+\frac{4}{5}\times 15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1\\6\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=1,y=6$$

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.

$$4x+y=10,3x+2y=15$$

To make 4x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 4.

$$3\times 4x+3y=3\times 10,4\times 3x+4\times 2y=4\times 15$$

Simplify.

$$12x+3y=30,12x+8y=60$$

Subtract 12x+8y=60 from 12x+3y=30 by subtracting like terms on each side of the equal sign.

$$12x-12x+3y-8y=30-60$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$3y-8y=30-60$$

Add 3y to -8y.

$$-5y=30-60$$

Add 30 to -60.

$$-5y=-30$$

Divide both sides by -5.

$$y=6$$

Substitute 6 for y in 3x+2y=15. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x+2\times 6=15$$

Multiply 2 times 6.

$$3x+12=15$$

Subtract 12 from both sides of the equation.

$$3x=3$$

Divide both sides by 3.

$$x=1$$

The system is now solved.

$$x=1,y=6$$