Solve {l}{3-2(y-4x)=4(1-x)-19}{-3(2x-y+1)=-2y+26}

$$\left\{ \begin{array} { l } { 3 - 2 ( y - 4 x ) = 4 ( 1 - x ) - 19 } \\ { - 3 ( 2 x - y + 1 ) = - 2 y + 26 } \end{array} \right.$$

Solve for y, x

$$x=-\frac{2}{3}\approx -0.666666667$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply -2 by y-4x.

$$3-2y+8x=4\left(1-x\right)-19$$

Use the distributive property to multiply 4 by 1-x.

$$3-2y+8x=4-4x-19$$

Subtract 19 from 4 to get -15.

$$3-2y+8x=-15-4x$$

Add 4x to both sides.

$$3-2y+8x+4x=-15$$

Combine 8x and 4x to get 12x.

$$3-2y+12x=-15$$

Subtract 3 from both sides.

$$-2y+12x=-15-3$$

Subtract 3 from -15 to get -18.

$$-2y+12x=-18$$

Consider the second equation. Use the distributive property to multiply -3 by 2x-y+1.

$$-6x+3y-3=-2y+26$$

Add 2y to both sides.

$$-6x+3y-3+2y=26$$

Combine 3y and 2y to get 5y.

$$-6x+5y-3=26$$

Add 3 to both sides.

$$-6x+5y=26+3$$

Add 26 and 3 to get 29.

$$-6x+5y=29$$

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.

$$-2y+12x=-18,5y-6x=29$$

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

$$-2y+12x=-18$$

Subtract 12x from both sides of the equation.

$$-2y=-12x-18$$

Divide both sides by -2.

$$y=-\frac{1}{2}\left(-12x-18\right)$$

Multiply -\frac{1}{2} times -12x-18.

$$y=6x+9$$

Substitute 6x+9 for y in the other equation, 5y-6x=29.

$$5\left(6x+9\right)-6x=29$$

Multiply 5 times 6x+9.

$$30x+45-6x=29$$

Add 30x to -6x.

$$24x+45=29$$

Subtract 45 from both sides of the equation.

$$24x=-16$$

Divide both sides by 24.

$$x=-\frac{2}{3}$$

Substitute -\frac{2}{3} for x in y=6x+9. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=6\left(-\frac{2}{3}\right)+9$$

Multiply 6 times -\frac{2}{3}.

$$y=-4+9$$

Add 9 to -4.

$$y=5$$

The system is now solved.

$$y=5,x=-\frac{2}{3}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply -2 by y-4x.

$$3-2y+8x=4\left(1-x\right)-19$$

Use the distributive property to multiply 4 by 1-x.

$$3-2y+8x=4-4x-19$$

Subtract 19 from 4 to get -15.

$$3-2y+8x=-15-4x$$

Add 4x to both sides.

$$3-2y+8x+4x=-15$$

Combine 8x and 4x to get 12x.

$$3-2y+12x=-15$$

Subtract 3 from both sides.

$$-2y+12x=-15-3$$

Subtract 3 from -15 to get -18.

$$-2y+12x=-18$$

Consider the second equation. Use the distributive property to multiply -3 by 2x-y+1.

$$-6x+3y-3=-2y+26$$

Add 2y to both sides.

$$-6x+3y-3+2y=26$$

Combine 3y and 2y to get 5y.

$$-6x+5y-3=26$$

Add 3 to both sides.

$$-6x+5y=26+3$$

Add 26 and 3 to get 29.

$$-6x+5y=29$$

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

$$-2y+12x=-18,5y-6x=29$$

Write the equations in matrix form.

$$\left(\begin{matrix}-2&12\\5&-6\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-18\\29\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-2&12\\5&-6\end{matrix}\right).

$$inverse(\left(\begin{matrix}-2&12\\5&-6\end{matrix}\right))\left(\begin{matrix}-2&12\\5&-6\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-2&12\\5&-6\end{matrix}\right))\left(\begin{matrix}-18\\29\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-2&12\\5&-6\end{matrix}\right))\left(\begin{matrix}-18\\29\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}-2&12\\5&-6\end{matrix}\right))\left(\begin{matrix}-18\\29\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{-2\left(-6\right)-12\times 5}&-\frac{12}{-2\left(-6\right)-12\times 5}\\-\frac{5}{-2\left(-6\right)-12\times 5}&-\frac{2}{-2\left(-6\right)-12\times 5}\end{matrix}\right)\left(\begin{matrix}-18\\29\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}&\frac{1}{4}\\\frac{5}{48}&\frac{1}{24}\end{matrix}\right)\left(\begin{matrix}-18\\29\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}\left(-18\right)+\frac{1}{4}\times 29\\\frac{5}{48}\left(-18\right)+\frac{1}{24}\times 29\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}5\\-\frac{2}{3}\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=5,x=-\frac{2}{3}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply -2 by y-4x.

$$3-2y+8x=4\left(1-x\right)-19$$

Use the distributive property to multiply 4 by 1-x.

$$3-2y+8x=4-4x-19$$

Subtract 19 from 4 to get -15.

$$3-2y+8x=-15-4x$$

Add 4x to both sides.

$$3-2y+8x+4x=-15$$

Combine 8x and 4x to get 12x.

$$3-2y+12x=-15$$

Subtract 3 from both sides.

$$-2y+12x=-15-3$$

Subtract 3 from -15 to get -18.

$$-2y+12x=-18$$

Consider the second equation. Use the distributive property to multiply -3 by 2x-y+1.

$$-6x+3y-3=-2y+26$$

Add 2y to both sides.

$$-6x+3y-3+2y=26$$

Combine 3y and 2y to get 5y.

$$-6x+5y-3=26$$

Add 3 to both sides.

$$-6x+5y=26+3$$

Add 26 and 3 to get 29.

$$-6x+5y=29$$

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.

$$-2y+12x=-18,5y-6x=29$$

To make -2y and 5y equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by -2.

$$5\left(-2\right)y+5\times 12x=5\left(-18\right),-2\times 5y-2\left(-6\right)x=-2\times 29$$

Simplify.

$$-10y+60x=-90,-10y+12x=-58$$

Subtract -10y+12x=-58 from -10y+60x=-90 by subtracting like terms on each side of the equal sign.

$$-10y+10y+60x-12x=-90+58$$

Add -10y to 10y. Terms -10y and 10y cancel out, leaving an equation with only one variable that can be solved.

$$60x-12x=-90+58$$

Add 60x to -12x.

$$48x=-90+58$$

Add -90 to 58.

$$48x=-32$$

Divide both sides by 48.

$$x=-\frac{2}{3}$$

Substitute -\frac{2}{3} for x in 5y-6x=29. Because the resulting equation contains only one variable, you can solve for y directly.

$$5y-6\left(-\frac{2}{3}\right)=29$$

Multiply -6 times -\frac{2}{3}.

$$5y+4=29$$

Subtract 4 from both sides of the equation.

$$5y=25$$

Divide both sides by 5.

$$y=5$$

The system is now solved.

$$y=5,x=-\frac{2}{3}$$