When c = 1 and d = 4,\(\frac{(cd)^{2}}{c+d}\) =
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Solution
Substitute 1 for each instance of c and substitute 4 for each instance of d in the expression: \(\frac{((1)(4))^{2}}{(1)+(4)}=\frac{4^{2}}{5}=\frac{16}{5}\).
When a = 3, (4a2)(3b3 + a) – b3 =
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Solution
Substitute 3 for each instance of a in the expression: (4(3)2)(3b3 + (3)) – b3 = (4)(9)(3b3 + 3) –b3 = 36(3b3 + 3) – b3 = 108b3 + 108 – b3 = 107b3 + 108.
When p = 6,\(\frac{4p(p+r)}{pr}\)=
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Solution
Substitute 6 for each instance of a in the expression:\(\frac{4(6)((6)+r)}{6r}=\frac{24(6+r)}{6r}=\frac{4(6+r)}{r}=\frac{24+4r}{r}\). The expression cannot be simplified any further.
When x = –2,\(\frac{y^{2}}{x^{2}}+\frac{y}{-2x}\) =
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Solution
Substitute –2 for each instance of x in the expression:\(\frac{(y^{2})}{(-2)^{2}}+\frac{y}{(-2(-2))}=\frac{y^{2}}{4}+\frac{y}{4}=\frac{(y^{2}+y)}{4}\).
When a = –2,\(\frac{7a}{a^{2}+a}\) =
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Solution
Substitute –2 for each instance of a in the expression:\(\frac{7(-2)}{((-2)^{2}+(-2))}=\frac{-14}{(4-2)}=-\frac{14}{2}=-7\).
When x = 3, 2x2 – 5x + 3 =
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Solution
Substitute 3 for each instance of x in the expression: 2(3)2 – 5(3) + 3 = 2(9) – 15 + 3 = 18 – 15 + 3 = 6.
(2x2)(4y2) + 6x2y2 =
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Solution
Multiply 2x2 and 4y2 by multiplying the coefficients of the terms: (2x2)(4y2) = 8 x2y2. 8x2y2 and 6x2y2 have like bases, so they can be added. Add the coefficients: 8x2y2 + 6x2y2 = 14x2y2.
\(\frac{(5a+7b)b}{(b+2b)}\) =
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Solution
The terms 5a and 7b have unlike bases; they cannot be combined any further. Add the terms in the denominator; b + 2b = 3b. Divide the b term in the numerator by the 3b in the denominator;\(\frac{b}{3b}=\frac{1}{3}.(5a+7b)(\frac{1}{3})=\frac{5a+7b}{3}\).
\(\frac{(3a)(4a)}{6(6a^{2})}\)=
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Solution
Multiply the coefficients of the terms in the numerator, and add the exponents of the bases:(3a)(4a) = 12a2. Do the same with the terms in the denominator: [6(6a2)] = 36a2. Finally, divide the numerator by the denominator. Divide the coefficients of the terms and subtract the exponents of the bases: \(\frac{\left ( 12a^{2} \right )}{\left (36a^{2} \right )}=\frac{1}{3}\)
9a + 12a2 – 5a =
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Solution
Subtract the like terms by subtracting the coefficients of the terms: 9a – 5a = 4a. 4a and 12a2 are not like terms, so they cannot be combined any further; 9a + 12a2 – 5a = 12a2 + 4a.