Solve {l}{x-4x+1/9=2y-5/3}{y-3y+2/7=x+18/10}

$$\left\{ \begin{array} { l } { x - \frac { 4 x + 1 } { 9 } = \frac { 2 y - 5 } { 3 } } \\ { y - \frac { 3 y + 2 } { 7 } = \frac { x + 18 } { 10 } } \end{array} \right.$$

Solve for x, y

$$x=2$$

$$y=4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 9, the least common multiple of 9,3.

$$9x-\left(4x+1\right)=3\left(2y-5\right)$$

To find the opposite of 4x+1, find the opposite of each term.

$$9x-4x-1=3\left(2y-5\right)$$

Combine 9x and -4x to get 5x.

$$5x-1=3\left(2y-5\right)$$

Use the distributive property to multiply 3 by 2y-5.

$$5x-1=6y-15$$

Subtract 6y from both sides.

$$5x-1-6y=-15$$

Add 1 to both sides.

$$5x-6y=-15+1$$

Add -15 and 1 to get -14.

$$5x-6y=-14$$

Consider the second equation. Multiply both sides of the equation by 70, the least common multiple of 7,10.

$$70y-10\left(3y+2\right)=7\left(x+18\right)$$

Use the distributive property to multiply -10 by 3y+2.

$$70y-30y-20=7\left(x+18\right)$$

Combine 70y and -30y to get 40y.

$$40y-20=7\left(x+18\right)$$

Use the distributive property to multiply 7 by x+18.

$$40y-20=7x+126$$

Subtract 7x from both sides.

$$40y-20-7x=126$$

Add 20 to both sides.

$$40y-7x=126+20$$

Add 126 and 20 to get 146.

$$40y-7x=146$$

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.

$$5x-6y=-14,-7x+40y=146$$

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

$$5x-6y=-14$$

Add 6y to both sides of the equation.

$$5x=6y-14$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(6y-14\right)$$

Multiply \frac{1}{5} times 6y-14.

$$x=\frac{6}{5}y-\frac{14}{5}$$

Substitute \frac{6y-14}{5} for x in the other equation, -7x+40y=146.

$$-7\left(\frac{6}{5}y-\frac{14}{5}\right)+40y=146$$

Multiply -7 times \frac{6y-14}{5}.

$$-\frac{42}{5}y+\frac{98}{5}+40y=146$$

Add -\frac{42y}{5} to 40y.

$$\frac{158}{5}y+\frac{98}{5}=146$$

Subtract \frac{98}{5} from both sides of the equation.

$$\frac{158}{5}y=\frac{632}{5}$$

Divide both sides of the equation by \frac{158}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=4$$

Substitute 4 for y in x=\frac{6}{5}y-\frac{14}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{6}{5}\times 4-\frac{14}{5}$$

Multiply \frac{6}{5} times 4.

$$x=\frac{24-14}{5}$$

Add -\frac{14}{5} to \frac{24}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=2$$

The system is now solved.

$$x=2,y=4$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 9, the least common multiple of 9,3.

$$9x-\left(4x+1\right)=3\left(2y-5\right)$$

To find the opposite of 4x+1, find the opposite of each term.

$$9x-4x-1=3\left(2y-5\right)$$

Combine 9x and -4x to get 5x.

$$5x-1=3\left(2y-5\right)$$

Use the distributive property to multiply 3 by 2y-5.

$$5x-1=6y-15$$

Subtract 6y from both sides.

$$5x-1-6y=-15$$

Add 1 to both sides.

$$5x-6y=-15+1$$

Add -15 and 1 to get -14.

$$5x-6y=-14$$

Consider the second equation. Multiply both sides of the equation by 70, the least common multiple of 7,10.

$$70y-10\left(3y+2\right)=7\left(x+18\right)$$

Use the distributive property to multiply -10 by 3y+2.

$$70y-30y-20=7\left(x+18\right)$$

Combine 70y and -30y to get 40y.

$$40y-20=7\left(x+18\right)$$

Use the distributive property to multiply 7 by x+18.

$$40y-20=7x+126$$

Subtract 7x from both sides.

$$40y-20-7x=126$$

Add 20 to both sides.

$$40y-7x=126+20$$

Add 126 and 20 to get 146.

$$40y-7x=146$$

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

$$5x-6y=-14,-7x+40y=146$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-6\\-7&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-14\\146\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-6\\-7&40\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-6\\-7&40\end{matrix}\right))\left(\begin{matrix}5&-6\\-7&40\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-6\\-7&40\end{matrix}\right))\left(\begin{matrix}-14\\146\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-6\\-7&40\end{matrix}\right))\left(\begin{matrix}-14\\146\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-6\\-7&40\end{matrix}\right))\left(\begin{matrix}-14\\146\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40}{5\times 40-\left(-6\left(-7\right)\right)}&-\frac{-6}{5\times 40-\left(-6\left(-7\right)\right)}\\-\frac{-7}{5\times 40-\left(-6\left(-7\right)\right)}&\frac{5}{5\times 40-\left(-6\left(-7\right)\right)}\end{matrix}\right)\left(\begin{matrix}-14\\146\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{79}&\frac{3}{79}\\\frac{7}{158}&\frac{5}{158}\end{matrix}\right)\left(\begin{matrix}-14\\146\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{79}\left(-14\right)+\frac{3}{79}\times 146\\\frac{7}{158}\left(-14\right)+\frac{5}{158}\times 146\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2,y=4$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 9, the least common multiple of 9,3.

$$9x-\left(4x+1\right)=3\left(2y-5\right)$$

To find the opposite of 4x+1, find the opposite of each term.

$$9x-4x-1=3\left(2y-5\right)$$

Combine 9x and -4x to get 5x.

$$5x-1=3\left(2y-5\right)$$

Use the distributive property to multiply 3 by 2y-5.

$$5x-1=6y-15$$

Subtract 6y from both sides.

$$5x-1-6y=-15$$

Add 1 to both sides.

$$5x-6y=-15+1$$

Add -15 and 1 to get -14.

$$5x-6y=-14$$

Consider the second equation. Multiply both sides of the equation by 70, the least common multiple of 7,10.

$$70y-10\left(3y+2\right)=7\left(x+18\right)$$

Use the distributive property to multiply -10 by 3y+2.

$$70y-30y-20=7\left(x+18\right)$$

Combine 70y and -30y to get 40y.

$$40y-20=7\left(x+18\right)$$

Use the distributive property to multiply 7 by x+18.

$$40y-20=7x+126$$

Subtract 7x from both sides.

$$40y-20-7x=126$$

Add 20 to both sides.

$$40y-7x=126+20$$

Add 126 and 20 to get 146.

$$40y-7x=146$$

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.

$$5x-6y=-14,-7x+40y=146$$

To make 5x and -7x equal, multiply all terms on each side of the first equation by -7 and all terms on each side of the second by 5.

$$-7\times 5x-7\left(-6\right)y=-7\left(-14\right),5\left(-7\right)x+5\times 40y=5\times 146$$

Simplify.

$$-35x+42y=98,-35x+200y=730$$

Subtract -35x+200y=730 from -35x+42y=98 by subtracting like terms on each side of the equal sign.

$$-35x+35x+42y-200y=98-730$$

Add -35x to 35x. Terms -35x and 35x cancel out, leaving an equation with only one variable that can be solved.

$$42y-200y=98-730$$

Add 42y to -200y.

$$-158y=98-730$$

Add 98 to -730.

$$-158y=-632$$

Divide both sides by -158.

$$y=4$$

Substitute 4 for y in -7x+40y=146. Because the resulting equation contains only one variable, you can solve for x directly.

$$-7x+40\times 4=146$$

Multiply 40 times 4.

$$-7x+160=146$$

Subtract 160 from both sides of the equation.

$$-7x=-14$$

Divide both sides by -7.

$$x=2$$

The system is now solved.

$$x=2,y=4$$