Solve for x
x = \frac{49}{6} = 8\frac{1}{6} \approx 8.166666667
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\frac{15}{3}-\frac{1}{2}x+\frac{1}{3}=\frac{5}{4}
Convert 5 to fraction \frac{15}{3}.
\frac{15+1}{3}-\frac{1}{2}x=\frac{5}{4}
Since \frac{15}{3} and \frac{1}{3} have the same denominator, add them by adding their numerators.
\frac{16}{3}-\frac{1}{2}x=\frac{5}{4}
Add 15 and 1 to get 16.
-\frac{1}{2}x=\frac{5}{4}-\frac{16}{3}
Subtract \frac{16}{3} from both sides.
-\frac{1}{2}x=\frac{15}{12}-\frac{64}{12}
Least common multiple of 4 and 3 is 12. Convert \frac{5}{4} and \frac{16}{3} to fractions with denominator 12.
-\frac{1}{2}x=\frac{15-64}{12}
Since \frac{15}{12} and \frac{64}{12} have the same denominator, subtract them by subtracting their numerators.
-\frac{1}{2}x=-\frac{49}{12}
Subtract 64 from 15 to get -49.
x=-\frac{49}{12}\left(-2\right)
Multiply both sides by -2, the reciprocal of -\frac{1}{2}.
x=\frac{-49\left(-2\right)}{12}
Express -\frac{49}{12}\left(-2\right) as a single fraction.
x=\frac{98}{12}
Multiply -49 and -2 to get 98.
x=\frac{49}{6}
Reduce the fraction \frac{98}{12} to lowest terms by extracting and canceling out 2.
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y = 3x + 4
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Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}