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\frac{0.5x}{0.2}+\frac{-0.1}{0.2}-0.5=\frac{0.4x-0.6}{1.2}
Divide each term of 0.5x-0.1 by 0.2 to get \frac{0.5x}{0.2}+\frac{-0.1}{0.2}.
2.5x+\frac{-0.1}{0.2}-0.5=\frac{0.4x-0.6}{1.2}
Divide 0.5x by 0.2 to get 2.5x.
2.5x+\frac{-1}{2}-0.5=\frac{0.4x-0.6}{1.2}
Expand \frac{-0.1}{0.2} by multiplying both numerator and the denominator by 10.
2.5x-\frac{1}{2}-0.5=\frac{0.4x-0.6}{1.2}
Fraction \frac{-1}{2} can be rewritten as -\frac{1}{2} by extracting the negative sign.
2.5x-\frac{1}{2}-\frac{1}{2}=\frac{0.4x-0.6}{1.2}
Convert decimal number 0.5 to fraction \frac{5}{10}. Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.
2.5x+\frac{-1-1}{2}=\frac{0.4x-0.6}{1.2}
Since -\frac{1}{2} and \frac{1}{2} have the same denominator, subtract them by subtracting their numerators.
2.5x+\frac{-2}{2}=\frac{0.4x-0.6}{1.2}
Subtract 1 from -1 to get -2.
2.5x-1=\frac{0.4x-0.6}{1.2}
Divide -2 by 2 to get -1.
2.5x-1=\frac{0.4x}{1.2}+\frac{-0.6}{1.2}
Divide each term of 0.4x-0.6 by 1.2 to get \frac{0.4x}{1.2}+\frac{-0.6}{1.2}.
2.5x-1=\frac{1}{3}x+\frac{-0.6}{1.2}
Divide 0.4x by 1.2 to get \frac{1}{3}x.
2.5x-1=\frac{1}{3}x+\frac{-6}{12}
Expand \frac{-0.6}{1.2} by multiplying both numerator and the denominator by 10.
2.5x-1=\frac{1}{3}x-\frac{1}{2}
Reduce the fraction \frac{-6}{12} to lowest terms by extracting and canceling out 6.
2.5x-1-\frac{1}{3}x=-\frac{1}{2}
Subtract \frac{1}{3}x from both sides.
\frac{13}{6}x-1=-\frac{1}{2}
Combine 2.5x and -\frac{1}{3}x to get \frac{13}{6}x.
\frac{13}{6}x=-\frac{1}{2}+1
Add 1 to both sides.
\frac{13}{6}x=-\frac{1}{2}+\frac{2}{2}
Convert 1 to fraction \frac{2}{2}.
\frac{13}{6}x=\frac{-1+2}{2}
Since -\frac{1}{2} and \frac{2}{2} have the same denominator, add them by adding their numerators.
\frac{13}{6}x=\frac{1}{2}
Add -1 and 2 to get 1.
x=\frac{\frac{1}{2}}{\frac{13}{6}}
Divide both sides by \frac{13}{6}.
x=\frac{1}{2\times \frac{13}{6}}
Express \frac{\frac{1}{2}}{\frac{13}{6}} as a single fraction.
x=\frac{1}{\frac{13}{3}}
Multiply 2 and \frac{13}{6} to get \frac{13}{3}.