Simple Linear Equations

Solving Simple Linear Equations

(Part 3)

Notes: Pg 10-11

In conclusion, to solve equations involving fractional coefficients, we will MULTIPLY both sides of the equation by the reciprocal of the fraction.

Solve  $\frac{2}{5}x=4$  
   

Reciprocal of  $\frac{2}{5}$  is  $\frac{5}{2}$  

 $\frac{2}{5}x\times\frac{5}{2}=4\times\frac{5}{2}$ 

 $\therefore x=10$  

Ex 5.1 HW

Take out your YELLOW book.

We will be doing Q4 and Q8

Multiple Choice

Ex 5.1 HW Q4a

Solve  $$\frac{a}{11}=7$$  

• a = 7

• a = 77

• $$a=\frac{7}{11}$$  

• a = 18

Multiple Choice

Ex 5.1 HW Q4b

Solve  $$\frac{1}{15}b=-4$$  

• b = -60

• b = 11

• $$b=-\frac{4}{15}$$  

• $$b=-\frac{15}{4}$$  
  

Multiple Choice

Ex 5.1 HW Q4c

Solve  $$-\frac{c}{3}=13$$  

• -c = 39

• $$c=-\frac{13}{3}$$  

• $$c=-39$$  

• c = 16

Multiple Choice

Ex 5.1 HW Q4d

Solve  $$-\frac{1}{8}d=-6$$  

• $$d=-48$$  

• $$d=\frac{6}{8}$$  

• $$-d=-48$$  

• d = 48

Multiple Choice

Ex 5.1 HW Q8a

Solve  $$12+\frac{y}{5}=7$$  .

(Hint: minus 12 on both sides of the equation first.)

• $$y=-1$$  

• $$y=-5$$  

• $$y=-25$$  

• $$y=5$$  
  

Multiple Choice

 [2016/P1/Q3a] Solve

$$\frac{2x}{3}=5$$

• $$x=15$$  

• $$x=7.5$$  

• $$x=\frac{10}{3}$$  

• $$x=10$$  

Multiple Choice

[2013/P1/Q23a] Solve

$$\frac{1}{3}x-4=1$$  

• $$x=5$$  

• $$x=-9$$  

• $$x=15$$  

• $$x=\frac{1}{12}$$