Solve {l}{x+0.5y-3/x-5+7=0}{3y-10(x-1)/6+x-y/4+1=0}

$$\left. \begin{array} { l } { \frac { x + 0.5 y - 3 } { x - 5 } + 7 = 0 } \\ { \frac { 3 y - 10 ( x - 1 ) } { 6 } + \frac { x - y } { 4 } + 1 = 0 } \end{array} \right.$$

Solve for x, y

$$x=4$$

$$y=12$$

Steps Using Substitution

Consider the first equation. Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x+0.5y-3+\left(x-5\right)\times 7=0$$

Use the distributive property to multiply x-5 by 7.

$$x+0.5y-3+7x-35=0$$

Combine x and 7x to get 8x.

$$8x+0.5y-3-35=0$$

Subtract 35 from -3 to get -38.

$$8x+0.5y-38=0$$

Add 38 to both sides. Anything plus zero gives itself.

$$8x+0.5y=38$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 6,4.

$$2\left(3y-10\left(x-1\right)\right)+3\left(x-y\right)+12=0$$

Use the distributive property to multiply -10 by x-1.

$$2\left(3y-10x+10\right)+3\left(x-y\right)+12=0$$

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

$$6y-20x+20+3\left(x-y\right)+12=0$$

Use the distributive property to multiply 3 by x-y.

$$6y-20x+20+3x-3y+12=0$$

Combine -20x and 3x to get -17x.

$$6y-17x+20-3y+12=0$$

Combine 6y and -3y to get 3y.

$$3y-17x+20+12=0$$

Add 20 and 12 to get 32.

$$3y-17x+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$3y-17x=-32$$

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.

$$8x+0.5y=38,-17x+3y=-32$$

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

$$8x+0.5y=38$$

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

$$8x=-0.5y+38$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-0.5y+38\right)$$

Multiply \frac{1}{8} times -\frac{y}{2}+38.

$$x=-\frac{1}{16}y+\frac{19}{4}$$

Substitute -\frac{y}{16}+\frac{19}{4} for x in the other equation, -17x+3y=-32.

$$-17\left(-\frac{1}{16}y+\frac{19}{4}\right)+3y=-32$$

Multiply -17 times -\frac{y}{16}+\frac{19}{4}.

$$\frac{17}{16}y-\frac{323}{4}+3y=-32$$

Add \frac{17y}{16} to 3y.

$$\frac{65}{16}y-\frac{323}{4}=-32$$

Add \frac{323}{4} to both sides of the equation.

$$\frac{65}{16}y=\frac{195}{4}$$

Divide both sides of the equation by \frac{65}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=12$$

Substitute 12 for y in x=-\frac{1}{16}y+\frac{19}{4}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{16}\times 12+\frac{19}{4}$$

Multiply -\frac{1}{16} times 12.

$$x=\frac{-3+19}{4}$$

Add \frac{19}{4} to -\frac{3}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=4$$

The system is now solved.

$$x=4,y=12$$

Steps Using Matrices

Consider the first equation. Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x+0.5y-3+\left(x-5\right)\times 7=0$$

Use the distributive property to multiply x-5 by 7.

$$x+0.5y-3+7x-35=0$$

Combine x and 7x to get 8x.

$$8x+0.5y-3-35=0$$

Subtract 35 from -3 to get -38.

$$8x+0.5y-38=0$$

Add 38 to both sides. Anything plus zero gives itself.

$$8x+0.5y=38$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 6,4.

$$2\left(3y-10\left(x-1\right)\right)+3\left(x-y\right)+12=0$$

Use the distributive property to multiply -10 by x-1.

$$2\left(3y-10x+10\right)+3\left(x-y\right)+12=0$$

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

$$6y-20x+20+3\left(x-y\right)+12=0$$

Use the distributive property to multiply 3 by x-y.

$$6y-20x+20+3x-3y+12=0$$

Combine -20x and 3x to get -17x.

$$6y-17x+20-3y+12=0$$

Combine 6y and -3y to get 3y.

$$3y-17x+20+12=0$$

Add 20 and 12 to get 32.

$$3y-17x+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$3y-17x=-32$$

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

$$8x+0.5y=38,-17x+3y=-32$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}38\\-32\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right))\left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right))\left(\begin{matrix}38\\-32\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right))\left(\begin{matrix}38\\-32\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&0.5\\-17&3\end{matrix}\right))\left(\begin{matrix}38\\-32\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{8\times 3-0.5\left(-17\right)}&-\frac{0.5}{8\times 3-0.5\left(-17\right)}\\-\frac{-17}{8\times 3-0.5\left(-17\right)}&\frac{8}{8\times 3-0.5\left(-17\right)}\end{matrix}\right)\left(\begin{matrix}38\\-32\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{65}&-\frac{1}{65}\\\frac{34}{65}&\frac{16}{65}\end{matrix}\right)\left(\begin{matrix}38\\-32\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{65}\times 38-\frac{1}{65}\left(-32\right)\\\frac{34}{65}\times 38+\frac{16}{65}\left(-32\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\12\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=4,y=12$$

Steps Using Elimination

Consider the first equation. Variable x cannot be equal to 5 since division by zero is not defined. Multiply both sides of the equation by x-5.

$$x+0.5y-3+\left(x-5\right)\times 7=0$$

Use the distributive property to multiply x-5 by 7.

$$x+0.5y-3+7x-35=0$$

Combine x and 7x to get 8x.

$$8x+0.5y-3-35=0$$

Subtract 35 from -3 to get -38.

$$8x+0.5y-38=0$$

Add 38 to both sides. Anything plus zero gives itself.

$$8x+0.5y=38$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 6,4.

$$2\left(3y-10\left(x-1\right)\right)+3\left(x-y\right)+12=0$$

Use the distributive property to multiply -10 by x-1.

$$2\left(3y-10x+10\right)+3\left(x-y\right)+12=0$$

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

$$6y-20x+20+3\left(x-y\right)+12=0$$

Use the distributive property to multiply 3 by x-y.

$$6y-20x+20+3x-3y+12=0$$

Combine -20x and 3x to get -17x.

$$6y-17x+20-3y+12=0$$

Combine 6y and -3y to get 3y.

$$3y-17x+20+12=0$$

Add 20 and 12 to get 32.

$$3y-17x+32=0$$

Subtract 32 from both sides. Anything subtracted from zero gives its negation.

$$3y-17x=-32$$

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.

$$8x+0.5y=38,-17x+3y=-32$$

To make 8x and -17x equal, multiply all terms on each side of the first equation by -17 and all terms on each side of the second by 8.

$$-17\times 8x-17\times 0.5y=-17\times 38,8\left(-17\right)x+8\times 3y=8\left(-32\right)$$

Simplify.

$$-136x-8.5y=-646,-136x+24y=-256$$

Subtract -136x+24y=-256 from -136x-8.5y=-646 by subtracting like terms on each side of the equal sign.

$$-136x+136x-8.5y-24y=-646+256$$

Add -136x to 136x. Terms -136x and 136x cancel out, leaving an equation with only one variable that can be solved.

$$-8.5y-24y=-646+256$$

Add -\frac{17y}{2} to -24y.

$$-32.5y=-646+256$$

Add -646 to 256.

$$-32.5y=-390$$

Divide both sides of the equation by -32.5, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=12$$

Substitute 12 for y in -17x+3y=-32. Because the resulting equation contains only one variable, you can solve for x directly.

$$-17x+3\times 12=-32$$

Multiply 3 times 12.

$$-17x+36=-32$$

Subtract 36 from both sides of the equation.

$$-17x=-68$$

Divide both sides by -17.

$$x=4$$

The system is now solved.

$$x=4,y=12$$