MAT10:1.1.4 Solving equations involving brackets

Solve the equation 2(x + 3) = 8

3
1
7
-1

Solve the equation 3(2x – 1) =-21

-3
-25/2
4
-8

Solve the equation -2(2x – 3) = 14

19/2
5
-2
11

Solve the equation -(3 – 4x) = -1

1/2
-1/2
1
-1

Solve the equation 4(3 + 2x) = -20

-15
-1
-9/2
-4

Solve the equation -2(5 – 2a)=-12

9/2
-1/2
1/2
5/2

Solve the equation -3(2 + x)=0

-2
1
5
0

Solve the equation 3(2x – 1) = 27

25/2
5
4
13

Solve the equation 4(1 – 2x) = -8

5/2
3/2
11/2
1/2

Solve the equation – (2 -3x) = -8

2
4/3
-10/3
-2

Solve the equation 2(a+2)=3(a-1)

a=-7
a=7
a=7/5
a=1

Solve the equation 5 – 2(p + 3) = 3

p=-2/3
p=-7/2
p=-4
p=-3

S0lve the equation 3x – (4 – 3x) = -10

x=-1
x=-7/3
x=2
x=undefined

Solve the equation 2(m – 3) – 3(m + 2) = -14

m=-26/5
m=4
m=14
m=2

Solve the equation -(3 – 2x) – 2(1 +3 x) = 7

x=-8/7
x=3/2
x=2
x=-3

The length of a rectangle A is (x + 3) cm and its width is 5 cm. The length of another rectangle B is (2x – 1) cm and its width is 3 cm. Given the sum of the areas of the two rectangles is 23 cm², find the length of rectangle A.

1 cm
5/8 cm
4 cm
4.6 cm

The length of a rectangle P is (2x + 1) cm and its width is x cm. A rectangle Q of length (x + 1) cm and width x cm is cut from the rectangle P. Given the area of the remaining shape is 225 cm², find the value of x.

15
-15
112.5
56.25

A triangle has base length 8 cm and height (2x + 1) cm. Find the height of the triangle if its area is 28 cm².

7 cm
3 cm
3.5 cm
4 cm

The dimensions of a cuboid are 3 cm by 4 cm by (4 – 3x) cm. Find the value of x, if the volume of the cuboid is 120 cm³.

2
24
-24
-2

The length of a cube is (2x + 1) cm and its volume is 27 cm³. Find the total surface area of the cuboid.

54 cm²
96 cm²
125 cm²
36 cm²

x = 4
x = 3
x = -1
x = 8/3