Y9 Math Revision (Algebra) 1

Solve for b.

9b=18

Solve for x.

x + 17 = 4

Solve for n.

-6n + 6 = 60

Solve for x.

$$\frac{x}{5}-10=-10$$  

Solve for r.

$$-6r-9=111$$  

Solve for x.

$$8\left(1+x\right)=-8$$  

Solve for a.

-a + a = 0

Solve for x.

$$3\left(1+7x\right)-7=-88$$  

Solve for n.

$$n+14\ =\ 5n-2$$  

Solve for r.

$$-26+4r=-2\left(3-4r\right)$$  

Solve for x.

$$\frac{10}{5}=\frac{x}{2}$$  

Solve for x.

$$\frac{9}{8}=\frac{10}{\left(x-3\right)}$$  

Expand and/or simplify this:

$$\frac{1}{4}\left(p+2q\right)-\frac{1}{5}\left(2p+q\right)$$   $$\$$  

Express the following as a fraction in its simplest form

$$3+\frac{2x-1}{3}+\frac{5x-1}{4}$$  

Express the following as a fraction in its simplest form

$$\frac{2\left(3x-2y\right)}{5}-\frac{3\left(y+2x\right)}{10}+1$$  

If $$s=3t+\frac{1}{2}av$$  , when a= 3, v = -2, and s= 7

find the value of t

Solve the following inequalities.

-8y + 3y < -16

During the Singapore 2010 Youth Olympics Games, there were (3x + 8y+ 5) athletes living in a residential hall. If (2x - y + 18) athletes left for their respective competitions, find the number of athletes who remained in the hall.

Timothy is x years old. Laura is half as old as Timothy while Mike is now twice as old as Timothy was 3 years ago. Find the sum of their present ages.

Two consecutive numbers are such that the sum of the larger number and twice the smaller number is 88. Find the larger number.

Thomas cycles the first part of a 102-km journey at an average speed of 36 km/h and walks the remaining distance at an average speed of 8 km/h. If he takes 4 hours for his entire journey, find the distance that he walks.