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example 2. evaluate $\frac{2}{3}8(a - b)^2+3b$ if $a = 5$ and $b = 2$.
$\frac{2}{3}8(a - b)^2+3b=\frac{2}{3}8(5 - 2)^2+3cdot2$ substitute 5 for a and 2 for b.
$=\frac{2}{3}8(3)^2+3cdot2$ evaluate inner parentheses first.
$=\frac{2}{3}8cdot9 + 3cdot2$ compute values with exponents.
$=\frac{2}{3}72 + 6$ multiply (left to right).
$=\frac{2}{3}78$ add.
$=\frac{2}{3}\frac{78}{1}=52$ multiply and simplify.
exercise 1. evaluating expressions.
evaluate each expression using $a = 3,b = 7,c=-2,d=\frac{1}{2}$, and $f = 0.3$. show your work.
- $a + b + c=$
- $(a + c)^2 - f=$
- $3ab-6bc=$
- $5c+bf=$
- $\frac{6ad}{c}=$
- $\frac{1}{3}(b - c)^2 - a=$ squared
- $2(a + b)+c=$
- $a+(fcdot a)=$
- $db - a + c^2=$
- $\frac{10f^2}{2d}=$
simplifying expressions
an expression in simplest form has no like terms, no parentheses, no powers of powers, and only positive exponents.
combining like terms
combine like terms by adding/subtracting coefficients. never change variable exponents when combining like terms. (commutative and other properties of real numbers are reviewed in appendix a, p. 27.)
example 3. simplify. $4xy^3+(-7xy^3)$
$4xy^3+(-7xy^3)=(4 + -7)xy^3=-3xy^3$ add like - term coefficients, 4 and - 7.
Step1: Substitute values for 1.
Substitute \(a = 3\), \(b = 7\), \(c=-2\) into \(a + b + c\).
\[3+7+( - 2)\]
Step2: Perform addition and subtraction.
\[3 + 7=10\], then \(10+( - 2)=8\)
Step3: Substitute values for 2.
Substitute \(a = 3\), \(c=-2\), \(f = 0.3\) into \((a + c)^{2}-f\). First find \(a + c\): \(3+( - 2)=1\), then \((a + c)^{2}=1^{2}=1\), and \((a + c)^{2}-f=1 - 0.3 = 0.7\)
Step4: Substitute values for 3.
Substitute \(a = 3\), \(b = 7\), \(c=-2\) into \(3ab-6bc\).
\[3\times3\times7-6\times7\times(-2)=63 + 84=147\]
Step5: Substitute values for 4.
Substitute \(c=-2\), \(b = 7\), \(f = 0.3\) into \(5c+bf\).
\[5\times(-2)+7\times0.3=-10 + 2.1=-7.9\]
Step6: Substitute values for 5.
Substitute \(a = 3\), \(d=\frac{1}{2}\), \(c=-2\) into \(\frac{6ad}{c}\).
\[6\times3\times\frac{1}{2}\div(-2)=9\div(-2)=-\frac{9}{2}=-4.5\]
Step7: Substitute values for 6.
Substitute \(a = 3\), \(b = 7\), \(c=-2\) into \(\frac{1}{3}[(b - c)^{2}-a]\). First find \(b - c\): \(7-( - 2)=9\), then \((b - c)^{2}=9^{2}=81\), and \(\frac{1}{3}[(b - c)^{2}-a]=\frac{1}{3}(81 - 3)=\frac{1}{3}\times78 = 26\)
Step8: Substitute values for 7.
Substitute \(a = 3\), \(b = 7\), \(c=-2\) into \(2(a + b)+c\). First find \(a + b\): \(3 + 7 = 10\), then \(2(a + b)=2\times10 = 20\), and \(2(a + b)+c=20+( - 2)=18\)
Step9: Substitute values for 8.
Substitute \(a = 3\), \(f = 0.3\) into \(a+(f\cdot a)\). \(f\cdot a=0.3\times3 = 0.9\), and \(a+(f\cdot a)=3+0.9 = 3.9\)
Step10: Substitute values for 9.
Substitute \(a = 3\), \(b = 7\), \(c=-2\), \(d=\frac{1}{2}\) into \(d[b - a + c]^{2}\). First find \(b - a + c\): \(7-3+( - 2)=2\), then \([b - a + c]^{2}=2^{2}=4\), and \(d[b - a + c]^{2}=\frac{1}{2}\times4 = 2\)
Step11: Substitute values for 10.
Substitute \(f = 0.3\), \(d=\frac{1}{2}\) into \(\frac{10f^{2}}{2d}\). First find \(f^{2}=0.3^{2}=0.09\), then \(10f^{2}=10\times0.09 = 0.9\), and \(\frac{10f^{2}}{2d}=\frac{0.9}{2\times\frac{1}{2}}=0.9\)
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- \(8\)
- \(0.7\)
- \(147\)
- \(-7.9\)
- \(-4.5\)
- \(26\)
- \(18\)
- \(3.9\)
- \(2\)
- \(0.9\)