Solve {l}{7/3x+y/2=4}{x+2/5=y+4/3}

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

Solve for x, y

$$x = \frac{162}{79} = 2\frac{4}{79} \approx 2.050632911$$

$$y = -\frac{124}{79} = -1\frac{45}{79} \approx -1.569620253$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$14x+3y=24$$

Consider the second equation. Multiply both sides of the equation by 15, the least common multiple of 5,3.

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

Use the distributive property to multiply 3 by x+2.

$$3x+6=5\left(y+4\right)$$

Use the distributive property to multiply 5 by y+4.

$$3x+6=5y+20$$

Subtract 5y from both sides.

$$3x+6-5y=20$$

Subtract 6 from both sides.

$$3x-5y=20-6$$

Subtract 6 from 20 to get 14.

$$3x-5y=14$$

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.

$$14x+3y=24,3x-5y=14$$

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

$$14x+3y=24$$

Subtract 3y from both sides of the equation.

$$14x=-3y+24$$

Divide both sides by 14.

$$x=\frac{1}{14}\left(-3y+24\right)$$

Multiply \frac{1}{14} times -3y+24.

$$x=-\frac{3}{14}y+\frac{12}{7}$$

Substitute -\frac{3y}{14}+\frac{12}{7} for x in the other equation, 3x-5y=14.

$$3\left(-\frac{3}{14}y+\frac{12}{7}\right)-5y=14$$

Multiply 3 times -\frac{3y}{14}+\frac{12}{7}.

$$-\frac{9}{14}y+\frac{36}{7}-5y=14$$

Add -\frac{9y}{14} to -5y.

$$-\frac{79}{14}y+\frac{36}{7}=14$$

Subtract \frac{36}{7} from both sides of the equation.

$$-\frac{79}{14}y=\frac{62}{7}$$

Divide both sides of the equation by -\frac{79}{14}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{124}{79}$$

Substitute -\frac{124}{79} for y in x=-\frac{3}{14}y+\frac{12}{7}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{14}\left(-\frac{124}{79}\right)+\frac{12}{7}$$

Multiply -\frac{3}{14} times -\frac{124}{79} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{186}{553}+\frac{12}{7}$$

Add \frac{12}{7} to \frac{186}{553} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{162}{79}$$

The system is now solved.

$$x=\frac{162}{79},y=-\frac{124}{79}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$14x+3y=24$$

Consider the second equation. Multiply both sides of the equation by 15, the least common multiple of 5,3.

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

Use the distributive property to multiply 3 by x+2.

$$3x+6=5\left(y+4\right)$$

Use the distributive property to multiply 5 by y+4.

$$3x+6=5y+20$$

Subtract 5y from both sides.

$$3x+6-5y=20$$

Subtract 6 from both sides.

$$3x-5y=20-6$$

Subtract 6 from 20 to get 14.

$$3x-5y=14$$

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

$$14x+3y=24,3x-5y=14$$

Write the equations in matrix form.

$$\left(\begin{matrix}14&3\\3&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}24\\14\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}14&3\\3&-5\end{matrix}\right).

$$inverse(\left(\begin{matrix}14&3\\3&-5\end{matrix}\right))\left(\begin{matrix}14&3\\3&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&3\\3&-5\end{matrix}\right))\left(\begin{matrix}24\\14\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&3\\3&-5\end{matrix}\right))\left(\begin{matrix}24\\14\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}14&3\\3&-5\end{matrix}\right))\left(\begin{matrix}24\\14\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{14\left(-5\right)-3\times 3}&-\frac{3}{14\left(-5\right)-3\times 3}\\-\frac{3}{14\left(-5\right)-3\times 3}&\frac{14}{14\left(-5\right)-3\times 3}\end{matrix}\right)\left(\begin{matrix}24\\14\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{79}&\frac{3}{79}\\\frac{3}{79}&-\frac{14}{79}\end{matrix}\right)\left(\begin{matrix}24\\14\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{79}\times 24+\frac{3}{79}\times 14\\\frac{3}{79}\times 24-\frac{14}{79}\times 14\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{162}{79}\\-\frac{124}{79}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{162}{79},y=-\frac{124}{79}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$14x+3y=24$$

Consider the second equation. Multiply both sides of the equation by 15, the least common multiple of 5,3.

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

Use the distributive property to multiply 3 by x+2.

$$3x+6=5\left(y+4\right)$$

Use the distributive property to multiply 5 by y+4.

$$3x+6=5y+20$$

Subtract 5y from both sides.

$$3x+6-5y=20$$

Subtract 6 from both sides.

$$3x-5y=20-6$$

Subtract 6 from 20 to get 14.

$$3x-5y=14$$

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.

$$14x+3y=24,3x-5y=14$$

To make 14x 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 14.

$$3\times 14x+3\times 3y=3\times 24,14\times 3x+14\left(-5\right)y=14\times 14$$

Simplify.

$$42x+9y=72,42x-70y=196$$

Subtract 42x-70y=196 from 42x+9y=72 by subtracting like terms on each side of the equal sign.

$$42x-42x+9y+70y=72-196$$

Add 42x to -42x. Terms 42x and -42x cancel out, leaving an equation with only one variable that can be solved.

$$9y+70y=72-196$$

Add 9y to 70y.

$$79y=72-196$$

Add 72 to -196.

$$79y=-124$$

Divide both sides by 79.

$$y=-\frac{124}{79}$$

Substitute -\frac{124}{79} for y in 3x-5y=14. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x-5\left(-\frac{124}{79}\right)=14$$

Multiply -5 times -\frac{124}{79}.

$$3x+\frac{620}{79}=14$$

Subtract \frac{620}{79} from both sides of the equation.

$$3x=\frac{486}{79}$$

Divide both sides by 3.

$$x=\frac{162}{79}$$

The system is now solved.

$$x=\frac{162}{79},y=-\frac{124}{79}$$