Beginner Level This section is for BEGINNERS. Skip to the GMAT Practice problems below if you already know the concept.

1Simplify $$8k^2 + 4k^2$$.
$$8k^2 + 4k^2 = 12k^2$$

2Simplify $$2y^2 + 6y^2 + 14y – 7y + 7$$.
$$2y^2 + 6y^2 + 14y – 7y + 7 \\ \, \\ = \underbrace{(2y^2 + 6y^2)}_{\text{Like-Terms}} + \underbrace{(14y – 7y)}_{\text{Like-Terms}} + 7 \\ \, \\ = 8y^2 + 7y + 7$$

3Simplify $$\dfrac{15a+5b}{3a+b}$$.
$$\require{cancel} \dfrac{15a+5b}{3a+b} \\ \, \\ =\dfrac{5(3a+b)}{3a+b} \\ \, \\ = \dfrac{5\cancel{({\color{red}{3a+b}})}}{\cancel{{\color{red}{3a+b}}}} \\ \, \\ = 5$$

4Simplify $$\dfrac{9x+27}{3\sqrt{3}}$$.
$$\require{cancel} \dfrac{9x+27}{3\sqrt{3}} \\ \, \\ =\dfrac{9(x+3)}{3\sqrt{3}} \\ \, \\ =\dfrac{9(x+3)}{3\sqrt{3}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \\ \, \\ =\dfrac{9\sqrt{3}(x+3)}{3\cdot3} \\ \, \\ =\dfrac{9\sqrt{3}(x+3)}{9} \\ \, \\ =\dfrac{\cancel{{\color{red}{9}}}\sqrt{3}(x+3)}{\cancel{{\color{red}{9}}}} \\ \, \\ = \sqrt{3}(x+3)$$

5Simplify $$\dfrac{8y+24}{\sqrt{8}}$$.
$$\require{cancel} \dfrac{8y+24}{\sqrt{8}} \\ \, \\ =\dfrac{8(y+3)}{\sqrt{8}} \\ \, \\ =\dfrac{8(y+3)}{\sqrt{8}}\cdot\dfrac{\sqrt{8}}{\sqrt{8}} \\ \, \\ =\dfrac{8\sqrt{8}(y+3)}{8} \\ \, \\ =\dfrac{\cancel{{\color{red}{8}}}\sqrt{8}(y+3)}{\cancel{{\color{red}{8}}}} \\ \, \\ =\sqrt{8}(y+3) \\ \, \\ =\sqrt{4\times2}(y+3) \\ \, \\ =2\sqrt{2}(y+3)$$

GMAT Practice

6If $$\dfrac{(12y^2+6y+36)y}{4y^2+2y+12}=1$$, what is the possible value of $$y$$?
A. $$3$$
B. $$1$$
C. $$\dfrac{1}{2}$$
D. $$\dfrac{1}{3}$$
E. $$\dfrac{1}{10}$$

Solution
Simplify the equation at numerator and denominator first by factorizing coefficients.
$$\require{cancel} \dfrac{(12y^2+6y+36)y}{4y^2+2y+12}=1 \\ \, \\ \dfrac{6(2y^2+y+6)y}{2(2y^2+y+6)}=1 \\ \, \\ \dfrac{6y\cancel{({\color{red}{2y^2+y+6}})}}{2({\cancel{{\color{red}{2y^2+y+6}}})}}=1 \\ \, \\ \dfrac{6y}{2}=1 \\ \, \\ 3y=1 \\ \, \\ y=\dfrac{1}{3} \\ \, \\$$

GMAT Practice

7If $$\dfrac{(x+y)^2 + (x-y)^2}{8}=2$$, what is the value of $$x^2+y^2$$?
A. 4
B. 6
C. 8
D. 16
E. 32

Solution
$$\require{cancel} \dfrac{(x+y)^2 + (x-y)^2}{8}=2 \\ \, \\ \dfrac{(x+y)^2 + (x-y)^2}{\cancel{{\color{red}{8}}}}\cdot\cancel{{\color{red}{8}}}=2\cdot8 \\ \, \\ (x+y)^2 + (x-y)^2 = 16 \\ \, \\ x^2\cancel{{\color{red}{+2xy}}}+y^2 + x^2 \cancel{{\color{red}{-2xy}}} + y^2 = 16 \\ \, \\ 2x^2 + 2y^2 = 16 \\ \, \\ 2(x^2+y^2) = 16 \\ \, \\ x^2 + y^2 = 8 \\ \, \\$$