Solve {l}{-12x+7=-7x+12}{|3x-11|+1/2y=|3x|-11+y}{(x,y)=(quad,quad)}

$$\left. \begin{array} { l } { - 12 x + 7 = - 7 x + 12 } \\ { | 3 x - 11 | + \frac { 1 } { 2 } y = | 3 x | - 11 + y } \\ { ( x , y ) = ( \quad , \quad ) } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=44$$

Solution Steps

Consider the first equation. Add 7x to both sides.

$$-12x+7+7x=12$$

Combine -12x and 7x to get -5x.

$$-5x+7=12$$

Subtract 7 from both sides.

$$-5x=12-7$$

Subtract 7 from 12 to get 5.

$$-5x=5$$

Divide both sides by -5.

$$x=\frac{5}{-5}$$

Divide 5 by -5 to get -1.

$$x=-1$$

Consider the second equation. Insert the known values of variables into the equation.

$$|3\left(-1\right)-11|+\frac{1}{2}y=|3\left(-1\right)|-11+y$$

Multiply 3 and -1 to get -3.

$$|-3-11|+\frac{1}{2}y=|3\left(-1\right)|-11+y$$

Subtract 11 from -3 to get -14.

$$|-14|+\frac{1}{2}y=|3\left(-1\right)|-11+y$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -14 is 14.

$$14+\frac{1}{2}y=|3\left(-1\right)|-11+y$$

Multiply 3 and -1 to get -3.

$$14+\frac{1}{2}y=|-3|-11+y$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of -3 is 3.

$$14+\frac{1}{2}y=3-11+y$$

Subtract 11 from 3 to get -8.

$$14+\frac{1}{2}y=-8+y$$

Subtract y from both sides.

$$14+\frac{1}{2}y-y=-8$$

Combine \frac{1}{2}y and -y to get -\frac{1}{2}y.

$$14-\frac{1}{2}y=-8$$

Subtract 14 from both sides.

$$-\frac{1}{2}y=-8-14$$

Subtract 14 from -8 to get -22.

$$-\frac{1}{2}y=-22$$

Multiply both sides by -2, the reciprocal of -\frac{1}{2}.

$$y=-22\left(-2\right)$$

Multiply -22 and -2 to get 44.

$$y=44$$

The system is now solved.

$$x=-1 y=44$$