Solve x/7+y/3=5;-x/14+3y=26

$$\frac { x } { 7 } + \frac { y } { 3 } = 5 ; - \frac { x } { 14 } + 3 y = 26$$

Solve for x, y

$$x=14$$

$$y=9$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 21, the least common multiple of 7,3.

$$3x+7y=105$$

Consider the second equation. Multiply both sides of the equation by 14.

$$-x+42y=364$$

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.

$$3x+7y=105,-x+42y=364$$

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

$$3x+7y=105$$

Subtract 7y from both sides of the equation.

$$3x=-7y+105$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-7y+105\right)$$

Multiply \frac{1}{3} times -7y+105.

$$x=-\frac{7}{3}y+35$$

Substitute -\frac{7y}{3}+35 for x in the other equation, -x+42y=364.

$$-\left(-\frac{7}{3}y+35\right)+42y=364$$

Multiply -1 times -\frac{7y}{3}+35.

$$\frac{7}{3}y-35+42y=364$$

Add \frac{7y}{3} to 42y.

$$\frac{133}{3}y-35=364$$

Add 35 to both sides of the equation.

$$\frac{133}{3}y=399$$

Divide both sides of the equation by \frac{133}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=9$$

Substitute 9 for y in x=-\frac{7}{3}y+35. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{7}{3}\times 9+35$$

Multiply -\frac{7}{3} times 9.

$$x=-21+35$$

Add 35 to -21.

$$x=14$$

The system is now solved.

$$x=14,y=9$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 21, the least common multiple of 7,3.

$$3x+7y=105$$

Consider the second equation. Multiply both sides of the equation by 14.

$$-x+42y=364$$

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

$$3x+7y=105,-x+42y=364$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&7\\-1&42\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}105\\364\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&7\\-1&42\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&7\\-1&42\end{matrix}\right))\left(\begin{matrix}3&7\\-1&42\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&7\\-1&42\end{matrix}\right))\left(\begin{matrix}105\\364\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&7\\-1&42\end{matrix}\right))\left(\begin{matrix}105\\364\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&7\\-1&42\end{matrix}\right))\left(\begin{matrix}105\\364\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{42}{3\times 42-7\left(-1\right)}&-\frac{7}{3\times 42-7\left(-1\right)}\\-\frac{-1}{3\times 42-7\left(-1\right)}&\frac{3}{3\times 42-7\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}105\\364\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{19}&-\frac{1}{19}\\\frac{1}{133}&\frac{3}{133}\end{matrix}\right)\left(\begin{matrix}105\\364\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{19}\times 105-\frac{1}{19}\times 364\\\frac{1}{133}\times 105+\frac{3}{133}\times 364\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\9\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=14,y=9$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 21, the least common multiple of 7,3.

$$3x+7y=105$$

Consider the second equation. Multiply both sides of the equation by 14.

$$-x+42y=364$$

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.

$$3x+7y=105,-x+42y=364$$

To make 3x and -x equal, multiply all terms on each side of the first equation by -1 and all terms on each side of the second by 3.

$$-3x-7y=-105,3\left(-1\right)x+3\times 42y=3\times 364$$

Simplify.

$$-3x-7y=-105,-3x+126y=1092$$

Subtract -3x+126y=1092 from -3x-7y=-105 by subtracting like terms on each side of the equal sign.

$$-3x+3x-7y-126y=-105-1092$$

Add -3x to 3x. Terms -3x and 3x cancel out, leaving an equation with only one variable that can be solved.

$$-7y-126y=-105-1092$$

Add -7y to -126y.

$$-133y=-105-1092$$

Add -105 to -1092.

$$-133y=-1197$$

Divide both sides by -133.

$$y=9$$

Substitute 9 for y in -x+42y=364. Because the resulting equation contains only one variable, you can solve for x directly.

$$-x+42\times 9=364$$

Multiply 42 times 9.

$$-x+378=364$$

Subtract 378 from both sides of the equation.

$$-x=-14$$

Divide both sides by -1.

$$x=14$$

The system is now solved.

$$x=14,y=9$$