2(x−3)=10{\displaystyle 2(x-3)=10} 2(x)−(2)(3)=10{\displaystyle 2(x)-(2)(3)=10} 2x−6=10{\displaystyle 2x-6=10}
2x−6=10{\displaystyle 2x-6=10}…. . (original problem) 2x−6(+6)=10(+6){\displaystyle 2x-6(+6)=10(+6)}…. . (Add 6 to both sides) 2x=16{\displaystyle 2x=16}…. . (Variable on left; constant on right)
2x=16{\displaystyle 2x=16}…. . (original problem) 2x/2=16/2{\displaystyle 2x/2=16/2}…. . (divide both sides by 2) x=8{\displaystyle x=8}…. . (solution)
Remember the basic rules of multiplying negatives: Neg. x Neg. = Pos. Neg. x Pos. = Neg. Consider the following example: −4(9−3x)=48{\displaystyle -4(9-3x)=48}…. . (original problem) −4(9)−(−4)(3x)=48{\displaystyle -4(9)-(-4)(3x)=48}…. . (distribute (-4) to each term) −36−(−12x)=48{\displaystyle -36-(-12x)=48}…. . (simplify the multiplication) −36+12x=48{\displaystyle -36+12x=48}…. . (notice that ‘minus -12’ becomes +12)
−36+12x=48{\displaystyle -36+12x=48}…. . (original problem) −36(+36)+12x=48+36{\displaystyle -36(+36)+12x=48+36}…. . (add 36 to each side) 12x=84{\displaystyle 12x=84}…. . (simplify the addition to isolate the variable)
12x=84{\displaystyle 12x=84}…. . (original problem) 12x/12=84/12{\displaystyle 12x/12=84/12}…. . (divide both sides by 12) x=7{\displaystyle x=7}…. . (solution)
For example, consider the problem, 4x−(x+2)=4{\displaystyle 4x-(x+2)=4}. To be certain that you distribute the negative properly, rewrite the problem to read: 4x+(−1)(x+2)=4{\displaystyle 4x+(-1)(x+2)=4} Then distribute the (-1) to the terms inside the parentheses as follows: 4x+(−1)(x+2)=4{\displaystyle 4x+(-1)(x+2)=4}…. . (revised problem) 4x−x−2=4{\displaystyle 4x-x-2=4}…. . (multiply (-1) times x and times 2) 3x−2=4{\displaystyle 3x-2=4}…. . (combine terms) 3x−2+2=4+2{\displaystyle 3x-2+2=4+2}…. . (add 2 to both sides) 3x=6{\displaystyle 3x=6}…. . (simplify terms) 3x/3=6/3{\displaystyle 3x/3=6/3}…. . (divide both sides by 3) x=2{\displaystyle x=2}…. . (solution)
Consider the example x−3=x3+16{\displaystyle x-3={\frac {x}{3}}+{\frac {1}{6}}}. The fractions in this problem are x3{\displaystyle {\frac {x}{3}}} and 16{\displaystyle {\frac {1}{6}}}.
x−3=x3+16{\displaystyle x-3={\frac {x}{3}}+{\frac {1}{6}}}…. . (original equation) (x−3)=(x3+16){\displaystyle (x-3)=({\frac {x}{3}}+{\frac {1}{6}})}…. . (insert parentheses) 6(x−3)=6(x3+16){\displaystyle 6(x-3)=6({\frac {x}{3}}+{\frac {1}{6}})}…. . (multiply both sides by LCM) 6x−6(3)=6(x3)+6(16){\displaystyle 6x-6(3)=6({\frac {x}{3}})+6({\frac {1}{6}})}…. . (distribute multiplication) 6x−18=2x+1{\displaystyle 6x-18=2x+1}…. . (simplify multiplication)
6x−18=2x+1{\displaystyle 6x-18=2x+1}…. . (simplified problem) 6x−2x−18=2x−2x+1{\displaystyle 6x-2x-18=2x-2x+1}…. . (subtract 2x from both sides) 4x−18=1{\displaystyle 4x-18=1}…. . (simplify subtraction) 4x−18+18=1+18{\displaystyle 4x-18+18=1+18}…. . (add 18 to both sides) 4x=19{\displaystyle 4x=19}…. . (simplify addition)
4x=19{\displaystyle 4x=19}…. . (revised problem) 4x/4=19/4{\displaystyle 4x/4=19/4}…. . (divide both sides by 4) x=194 or 434{\displaystyle x={\frac {19}{4}}{\text{ or }}4{\frac {3}{4}}}…. . (final solution)
4x+82=4{\displaystyle {\frac {4x+8}{2}}=4}. . . . . (original problem) 4x2+82=4{\displaystyle {\frac {4x}{2}}+{\frac {8}{2}}=4}. . . . . (distribute the denominator to each term of the numerator)
4x2+82=4{\displaystyle {\frac {4x}{2}}+{\frac {8}{2}}=4}. . . . . (revised problem) 2x+4=4{\displaystyle 2x+4=4}. . . . . (simplify the fractions)
2x+4=4{\displaystyle 2x+4=4}. . . . . (revised problem) 2x+4−4=4−4{\displaystyle 2x+4-4=4-4}. . . . . (subtract 4 from both sides) 2x=0{\displaystyle 2x=0}. . . . . (isolated x on one side)
2x=0{\displaystyle 2x=0}. . . . . (revised problem) 2x/2=0/2{\displaystyle 2x/2=0/2}. . . . . (divide both sides by 2) x=0{\displaystyle x=0}. . . . . (solution)
4x+82=4{\displaystyle {\frac {4x+8}{2}}=4}. . . . . (original problem) 2x+8=4{\displaystyle 2x+8=4}. . . . . (divide only 4x by 2 instead of the full numerator) 2x+8−8=4−8{\displaystyle 2x+8-8=4-8} 2x=−4{\displaystyle 2x=-4} x=−2{\displaystyle x=-2}. . . . . (incorrect solution)
Begin with the solution x=0: 4x+82=4{\displaystyle {\frac {4x+8}{2}}=4}. . . . . (original problem) 4(0)+82=4{\displaystyle {\frac {4(0)+8}{2}}=4}. . . . . (insert 0 for x) 0+82=4{\displaystyle {\frac {0+8}{2}}=4} 82=4{\displaystyle {\frac {8}{2}}=4} 4=4{\displaystyle 4=4}. . . . . (true statement. This is the correct solution. ) Try the “false” solution of x=-2: 4x+82=4{\displaystyle {\frac {4x+8}{2}}=4}. . . . . (original problem) 4(−2)+82=4{\displaystyle {\frac {4(-2)+8}{2}}=4}. . . . . (insert -2 for x) −8+82=4{\displaystyle {\frac {-8+8}{2}}=4} 02=4{\displaystyle {\frac {0}{2}}=4} 0=4{\displaystyle 0=4}. . . . . (incorrect statement. Therefore, x=-2 is false. )
