Solutions to Problem Set 1

Math 101-06

9-1-2017

1. Solve for x:

$$3 x + 17 = 7 + 21 x.$$

$$\eqalign{ 3 x + 17 & = 7 + 21 x \cr 3 x + 17 - 21 x - 17 & = 7 + 21 x - 21 x - 17 \cr -18 x & = -10 \cr \noalign{\vskip2pt} x & = \dfrac{5}{9} \quad\halmos \cr}$$


2. Solve for x:

$$-17 x + 2 = 6(5 x - 3).$$

$$\matrix{ -17 x + 2 & = & 6(5 x - 3) & \hbox{Multiply out parenthesized stuff} \cr -17 x + 2 & = & 30 x - 18 & \hbox{Move x-stuff to one side} \cr & & & \hbox{Move non-x-stuff to the other side} \cr -17 x + 2 - 30 x - 2 & = & 30 x - 18 - 30 x - 2 & \cr -47 x & = & -20 & \hbox{Divide to solve for x} \cr \noalign{\vskip2pt} x & = & \dfrac{20}{47} & \quad\halmos \cr}$$


3. Solve for x:

$$0.5 - 0.43 x = 0.3 x + 2$$

The decimals have one and two digits to the right of the decimal points. So I will clear the decimals by multiplying by $10^2 = 100$ , then solve for x:

$$\eqalign{ 100 \cdot (0.5 - 0.43 x) & = 100 \cdot (0.3 x + 2) \cr 100 \cdot 0.5 - 100 \cdot 0.43 x & = 100 \cdot 0.3 x + 100 \cdot 2 \cr 50 - 43 x & = 30 x + 200 \cr 50 - 43 x + 43 x - 200 & = 30 x + 200 + 43 x - 200 \cr -150 & = 73 x \cr \noalign{\vskip2pt} -\dfrac{150}{73} & = x \quad\halmos \cr}$$


4. Solve for x:

$$0.7 x - 3 = 0.37 (2 x - 6)$$

The decimals have one and two digits to the right of the decimal points. So I will clear the decimals by multiplying by $10^2 = 100$ , then solve for x:

$$\eqalign{ 0.7 x - 3 & = 0.37 (2 x - 6) \cr 100 \cdot 0.7 x - 100 \cdot 3 & = 100 \cdot 0.37 (2 x - 6) \cr 70 x - 300 & = 37 (2 x - 6) \cr 70 x - 300 & = 74 x - 222 \cr -4 x & = 78 \cr \noalign{\vskip2pt} x & = -\dfrac{78}{4} = -\dfrac{39}{2} \quad\halmos \cr}$$


5. Solve for x:

$$10 x + 7 = 5(2 x - 3)$$

$$\eqalign{ 10 x + 7 & = 5(2 x - 3) \cr 10 x + 7 & = 10 x - 15 \cr 7 & = -15 \cr}$$

The last equation is a contradiction. Hence, there are no solutions.


6. Solve for x:

$$\dfrac{5}{6} (7 x - 1) = \dfrac{1}{2} - \dfrac{3}{4} x$$

I will multiply to clear the denominators. The denominators of the fractions are 6, 2, and 4; the least common multiple of 6, 2, and 4 is 12. So I'll multiply both sides of the eqatuion by 12 and solve for x:

$$\eqalign{ \dfrac{5}{6} (7 x - 1) & = \dfrac{1}{2} - \dfrac{3}{4} x \cr \noalign{\vskip2pt} 12 \cdot \dfrac{5}{6} (7 x - 1) & = 12 \cdot \dfrac{1}{2} - 12 \cdot \dfrac{3}{4} x \cr \noalign{\vskip2pt} 10 (7 x - 1) & = 6 - 9 x \cr 70 x - 10 & = 6 - 9 x \cr 79 x & = 16 \cr \noalign{\vskip2pt} x & = \dfrac{16}{79} \quad\halmos \cr}$$


7. Solve for x:

$$\dfrac{2}{5} - \dfrac{1}{2} x = \dfrac{3}{4} (3 x - 1).$$

$$\eqalign{ \dfrac{2}{5} - \dfrac{1}{2} x & = \dfrac{3}{4} (3 x - 1) \cr \noalign{\vskip2pt} 20 \cdot \dfrac{2}{5} - 20 \cdot \dfrac{1}{2} x & = 20 \cdot \dfrac{3}{4} (3 x - 1) \cr \noalign{\vskip2pt} 8 - 10 x & = 15 (3 x - 1) \cr 8 - 10 x & = 45 x - 15 \cr 23 & = 55 x \cr \noalign{\vskip2pt} \dfrac{23}{55} & = x \quad\halmos \cr}$$


8. Solve for x:

$$8 x - 6 = 2(4 x - 3)$$

$$\eqalign{ 8 x - 6 & = 2(4 x - 3) \cr 8 x - 6 & = 8 x - 6 \cr 0 & = 0 \cr}$$

The last equation is an identity. Hence, the original equation is true for all x.

(You may write the answer as "all real numbers" or "$\real$ " or "$(-\infty, \infty)$ ".)


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