Solve {r}{36+7y=5x}{10x-42=4y}

$$\left. \begin{array} { r } { 36 + 7 y = 5 x } \\ { 10 x - 42 = 4 y } \end{array} \right.$$

Solve for y, x

$$x=3$$

$$y=-3$$

Steps Using Substitution

Consider the first equation. Subtract 5x from both sides.

$$36+7y-5x=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$7y-5x=-36$$

Consider the second equation. Subtract 4y from both sides.

$$10x-42-4y=0$$

Add 42 to both sides. Anything plus zero gives itself.

$$10x-4y=42$$

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.

$$7y-5x=-36,-4y+10x=42$$

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

$$7y-5x=-36$$

Add 5x to both sides of the equation.

$$7y=5x-36$$

Divide both sides by 7.

$$y=\frac{1}{7}\left(5x-36\right)$$

Multiply \frac{1}{7} times 5x-36.

$$y=\frac{5}{7}x-\frac{36}{7}$$

Substitute \frac{5x-36}{7} for y in the other equation, -4y+10x=42.

$$-4\left(\frac{5}{7}x-\frac{36}{7}\right)+10x=42$$

Multiply -4 times \frac{5x-36}{7}.

$$-\frac{20}{7}x+\frac{144}{7}+10x=42$$

Add -\frac{20x}{7} to 10x.

$$\frac{50}{7}x+\frac{144}{7}=42$$

Subtract \frac{144}{7} from both sides of the equation.

$$\frac{50}{7}x=\frac{150}{7}$$

Divide both sides of the equation by \frac{50}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=3$$

Substitute 3 for x in y=\frac{5}{7}x-\frac{36}{7}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{5}{7}\times 3-\frac{36}{7}$$

Multiply \frac{5}{7} times 3.

$$y=\frac{15-36}{7}$$

Add -\frac{36}{7} to \frac{15}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=-3$$

The system is now solved.

$$y=-3,x=3$$

Steps Using Matrices

Consider the first equation. Subtract 5x from both sides.

$$36+7y-5x=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$7y-5x=-36$$

Consider the second equation. Subtract 4y from both sides.

$$10x-42-4y=0$$

Add 42 to both sides. Anything plus zero gives itself.

$$10x-4y=42$$

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

$$7y-5x=-36,-4y+10x=42$$

Write the equations in matrix form.

$$\left(\begin{matrix}7&-5\\-4&10\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-36\\42\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-5\\-4&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}7&-5\\-4&10\end{matrix}\right))\left(\begin{matrix}7&-5\\-4&10\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-5\\-4&10\end{matrix}\right))\left(\begin{matrix}-36\\42\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-5\\-4&10\end{matrix}\right))\left(\begin{matrix}-36\\42\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}7&-5\\-4&10\end{matrix}\right))\left(\begin{matrix}-36\\42\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{10}{7\times 10-\left(-5\left(-4\right)\right)}&-\frac{-5}{7\times 10-\left(-5\left(-4\right)\right)}\\-\frac{-4}{7\times 10-\left(-5\left(-4\right)\right)}&\frac{7}{7\times 10-\left(-5\left(-4\right)\right)}\end{matrix}\right)\left(\begin{matrix}-36\\42\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}&\frac{1}{10}\\\frac{2}{25}&\frac{7}{50}\end{matrix}\right)\left(\begin{matrix}-36\\42\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\left(-36\right)+\frac{1}{10}\times 42\\\frac{2}{25}\left(-36\right)+\frac{7}{50}\times 42\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements y and x.

$$y=-3,x=3$$

Steps Using Elimination

Consider the first equation. Subtract 5x from both sides.

$$36+7y-5x=0$$

Subtract 36 from both sides. Anything subtracted from zero gives its negation.

$$7y-5x=-36$$

Consider the second equation. Subtract 4y from both sides.

$$10x-42-4y=0$$

Add 42 to both sides. Anything plus zero gives itself.

$$10x-4y=42$$

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.

$$7y-5x=-36,-4y+10x=42$$

To make 7y and -4y equal, multiply all terms on each side of the first equation by -4 and all terms on each side of the second by 7.

$$-4\times 7y-4\left(-5\right)x=-4\left(-36\right),7\left(-4\right)y+7\times 10x=7\times 42$$

Simplify.

$$-28y+20x=144,-28y+70x=294$$

Subtract -28y+70x=294 from -28y+20x=144 by subtracting like terms on each side of the equal sign.

$$-28y+28y+20x-70x=144-294$$

Add -28y to 28y. Terms -28y and 28y cancel out, leaving an equation with only one variable that can be solved.

$$20x-70x=144-294$$

Add 20x to -70x.

$$-50x=144-294$$

Add 144 to -294.

$$-50x=-150$$

Divide both sides by -50.

$$x=3$$

Substitute 3 for x in -4y+10x=42. Because the resulting equation contains only one variable, you can solve for y directly.

$$-4y+10\times 3=42$$

Multiply 10 times 3.

$$-4y+30=42$$

Subtract 30 from both sides of the equation.

$$-4y=12$$

Divide both sides by -4.

$$y=-3$$

The system is now solved.

$$y=-3,x=3$$