Solve {r}{4(4x+9y-3)=-4x-34}{10(6x-3y)+14=4(7x-3y+1)}

$$\left\{ \begin{array} { r } { 4 ( 4 x + 9 y - 3 ) = - 4 x - 34 } \\ { 10 ( 6 x - 3 y ) + 14 = 4 ( 7 x - 3 y + 1 ) } \end{array} \right.$$

Solve for x, y

$$x=-\frac{1}{2}=-0.5$$

$$y=-\frac{1}{3}\approx -0.333333333$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 4 by 4x+9y-3.

$$16x+36y-12=-4x-34$$

Add 4x to both sides.

$$16x+36y-12+4x=-34$$

Combine 16x and 4x to get 20x.

$$20x+36y-12=-34$$

Add 12 to both sides.

$$20x+36y=-34+12$$

Add -34 and 12 to get -22.

$$20x+36y=-22$$

Consider the second equation. Use the distributive property to multiply 10 by 6x-3y.

$$60x-30y+14=4\left(7x-3y+1\right)$$

Use the distributive property to multiply 4 by 7x-3y+1.

$$60x-30y+14=28x-12y+4$$

Subtract 28x from both sides.

$$60x-30y+14-28x=-12y+4$$

Combine 60x and -28x to get 32x.

$$32x-30y+14=-12y+4$$

Add 12y to both sides.

$$32x-30y+14+12y=4$$

Combine -30y and 12y to get -18y.

$$32x-18y+14=4$$

Subtract 14 from both sides.

$$32x-18y=4-14$$

Subtract 14 from 4 to get -10.

$$32x-18y=-10$$

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.

$$20x+36y=-22,32x-18y=-10$$

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

$$20x+36y=-22$$

Subtract 36y from both sides of the equation.

$$20x=-36y-22$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(-36y-22\right)$$

Multiply \frac{1}{20} times -36y-22.

$$x=-\frac{9}{5}y-\frac{11}{10}$$

Substitute -\frac{9y}{5}-\frac{11}{10} for x in the other equation, 32x-18y=-10.

$$32\left(-\frac{9}{5}y-\frac{11}{10}\right)-18y=-10$$

Multiply 32 times -\frac{9y}{5}-\frac{11}{10}.

$$-\frac{288}{5}y-\frac{176}{5}-18y=-10$$

Add -\frac{288y}{5} to -18y.

$$-\frac{378}{5}y-\frac{176}{5}=-10$$

Add \frac{176}{5} to both sides of the equation.

$$-\frac{378}{5}y=\frac{126}{5}$$

Divide both sides of the equation by -\frac{378}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{1}{3}$$

Substitute -\frac{1}{3} for y in x=-\frac{9}{5}y-\frac{11}{10}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{9}{5}\left(-\frac{1}{3}\right)-\frac{11}{10}$$

Multiply -\frac{9}{5} times -\frac{1}{3} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{3}{5}-\frac{11}{10}$$

Add -\frac{11}{10} to \frac{3}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{1}{2}$$

The system is now solved.

$$x=-\frac{1}{2},y=-\frac{1}{3}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 4 by 4x+9y-3.

$$16x+36y-12=-4x-34$$

Add 4x to both sides.

$$16x+36y-12+4x=-34$$

Combine 16x and 4x to get 20x.

$$20x+36y-12=-34$$

Add 12 to both sides.

$$20x+36y=-34+12$$

Add -34 and 12 to get -22.

$$20x+36y=-22$$

Consider the second equation. Use the distributive property to multiply 10 by 6x-3y.

$$60x-30y+14=4\left(7x-3y+1\right)$$

Use the distributive property to multiply 4 by 7x-3y+1.

$$60x-30y+14=28x-12y+4$$

Subtract 28x from both sides.

$$60x-30y+14-28x=-12y+4$$

Combine 60x and -28x to get 32x.

$$32x-30y+14=-12y+4$$

Add 12y to both sides.

$$32x-30y+14+12y=4$$

Combine -30y and 12y to get -18y.

$$32x-18y+14=4$$

Subtract 14 from both sides.

$$32x-18y=4-14$$

Subtract 14 from 4 to get -10.

$$32x-18y=-10$$

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

$$20x+36y=-22,32x-18y=-10$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&36\\32&-18\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-22\\-10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&36\\32&-18\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&36\\32&-18\end{matrix}\right))\left(\begin{matrix}20&36\\32&-18\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&36\\32&-18\end{matrix}\right))\left(\begin{matrix}-22\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&36\\32&-18\end{matrix}\right))\left(\begin{matrix}-22\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&36\\32&-18\end{matrix}\right))\left(\begin{matrix}-22\\-10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{18}{20\left(-18\right)-36\times 32}&-\frac{36}{20\left(-18\right)-36\times 32}\\-\frac{32}{20\left(-18\right)-36\times 32}&\frac{20}{20\left(-18\right)-36\times 32}\end{matrix}\right)\left(\begin{matrix}-22\\-10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{84}&\frac{1}{42}\\\frac{4}{189}&-\frac{5}{378}\end{matrix}\right)\left(\begin{matrix}-22\\-10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{84}\left(-22\right)+\frac{1}{42}\left(-10\right)\\\frac{4}{189}\left(-22\right)-\frac{5}{378}\left(-10\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\\-\frac{1}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{1}{2},y=-\frac{1}{3}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 4 by 4x+9y-3.

$$16x+36y-12=-4x-34$$

Add 4x to both sides.

$$16x+36y-12+4x=-34$$

Combine 16x and 4x to get 20x.

$$20x+36y-12=-34$$

Add 12 to both sides.

$$20x+36y=-34+12$$

Add -34 and 12 to get -22.

$$20x+36y=-22$$

Consider the second equation. Use the distributive property to multiply 10 by 6x-3y.

$$60x-30y+14=4\left(7x-3y+1\right)$$

Use the distributive property to multiply 4 by 7x-3y+1.

$$60x-30y+14=28x-12y+4$$

Subtract 28x from both sides.

$$60x-30y+14-28x=-12y+4$$

Combine 60x and -28x to get 32x.

$$32x-30y+14=-12y+4$$

Add 12y to both sides.

$$32x-30y+14+12y=4$$

Combine -30y and 12y to get -18y.

$$32x-18y+14=4$$

Subtract 14 from both sides.

$$32x-18y=4-14$$

Subtract 14 from 4 to get -10.

$$32x-18y=-10$$

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.

$$20x+36y=-22,32x-18y=-10$$

To make 20x and 32x equal, multiply all terms on each side of the first equation by 32 and all terms on each side of the second by 20.

$$32\times 20x+32\times 36y=32\left(-22\right),20\times 32x+20\left(-18\right)y=20\left(-10\right)$$

Simplify.

$$640x+1152y=-704,640x-360y=-200$$

Subtract 640x-360y=-200 from 640x+1152y=-704 by subtracting like terms on each side of the equal sign.

$$640x-640x+1152y+360y=-704+200$$

Add 640x to -640x. Terms 640x and -640x cancel out, leaving an equation with only one variable that can be solved.

$$1152y+360y=-704+200$$

Add 1152y to 360y.

$$1512y=-704+200$$

Add -704 to 200.

$$1512y=-504$$

Divide both sides by 1512.

$$y=-\frac{1}{3}$$

Substitute -\frac{1}{3} for y in 32x-18y=-10. Because the resulting equation contains only one variable, you can solve for x directly.

$$32x-18\left(-\frac{1}{3}\right)=-10$$

Multiply -18 times -\frac{1}{3}.

$$32x+6=-10$$

Subtract 6 from both sides of the equation.

$$32x=-16$$

Divide both sides by 32.

$$x=-\frac{1}{2}$$

The system is now solved.

$$x=-\frac{1}{2},y=-\frac{1}{3}$$