We have evaluated expressions before, but now we can also evaluate expressions with fractions. Remember, to evaluate an expression, we substitute the value of the variable into the expression and then simplify.
Example
Evaluate \(x+\frac{1}{3}\) when
\(x=-\frac{1}{3}\)
\(x=-\frac{3}{4}.\)
Solution
To evaluate \(x+\frac{1}{3}\) when \(x=-\frac{1}{3},\) substitute \(-\frac{1}{3}\) for \(x\) in the expression.
\(x+\frac{1}{3}\)
Simplify.
\(0\)
To evaluate \(x+\frac{1}{3}\) when \(x=-\frac{3}{4},\) we substitute \(-\frac{3}{4}\) for \(x\) in the expression.
\(x+\frac{1}{3}\)
Rewrite as equivalent fractions with the LCD, 12.
\(-\frac{3·3}{4·3}+\frac{1·4}{3·4}\)
Simplify the numerators and denominators.
\(-\frac{9}{12}+\frac{4}{12}\)
Add.
\(-\frac{5}{12}\)
Example
Evaluate \(y-\frac{5}{6}\) when \(y=-\frac{2}{3}.\)
Solution
We substitute \(-\frac{2}{3}\) for \(y\) in the expression.
\(y-\frac{5}{6}\)
Rewrite as equivalent fractions with the LCD, 6.
\(-\frac{4}{6}-\frac{5}{6}\)
Subtract.
\(-\frac{9}{6}\)
Simplify.
\(-\frac{3}{2}\)
Example
Evaluate \(2{x}^{2}y\) when \(x=\frac{1}{4}\) and \(y=-\frac{2}{3}.\)
Solution
Substitute the values into the expression. In \(2{x}^{2}y,\) the exponent applies only to \(x.\)
Simplify exponents first.
Multiply. The product will be negative.
Simplify.
Remove the common factors.
Simplify.
Example
Evaluate \(\frac{p+q}{r}\) when \(p=-4,q=-2,\) and \(r=8.\)
Solution
We substitute the values into the expression and simplify.
\(\frac{p+q}{r}\)
Add in the numerator first.
\(-\frac{6}{8}\)
Simplify.
\(-\frac{3}{4}\)
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