Algebra Questions with Solutions
and Explanations for Grade 9
Detailed solutions and full explanations to grade 9 algebra questions are presented.


Simplify the following algebraic expressions.
 – 6x + 5 + 12x 6

2(x – 9) + 6(x + 2) + 4x

3x^{2}
+ 12 + 9x – 20 + 6x^{2}
– x

(x + 2)(x + 4) + (x + 5)(x – 1)

1.2(x – 9) – 2.3(x + 4)

(x^{2}y)(xy^{2})

(x^{2}y^{2})(xy^{2})
Solution

Group like terms and simplify.
– 6x + 5 + 12x 6 = ( 6x + 12x) + (5 – 6)
= 6x – 1

Expand brackets.
2(x – 9) + 6(x + 2) + 4x = 2x – 18 – 6x + 12 + 4x
Group like terms and simplify.
= (2x – 6x + 4x) + ( 18 + 12) = – 6

Group like terms and simplify.
3x^{2}
+ 12 + 9x – 20 + 6x^{2}
– x
= (3x^{2}
+ 6x^{2}) + (9x – x) + (12 – 20)
= 9x^{2}
+ 8x – 8

Expand brackets.
(x + 2)(x + 4) + (x + 5)( x – 1)
= x^{2}
+ 4x + 2x + 8 – x^{2}
– x – 5x – 5
Group like terms.
= (x^{2}
– x^{2}) + (4x + 2x – x – 5x) + (8 – 5)
= 3

Expand and group.
1.2(x – 9) – 2.3(x + 4)
= 1.2x – 10.8 – 2.3x – 9.2
= 1.1x – 20

Rewrite as follows.
(x^{2}y)(xy^{2}) = (x^{2}
x)(y y^{2})
Use rules of exponential.
= x^{3}
y^{3}

Rewrite expression as follows.
(x^{2}y^{2})(xy^{2}) = (x^{2}
x)( y^{2}
y^{2})
Use rules of exponential.
= – x^{3}
y^{4}

Simplify the expressions.

(a b^{2})(a^{3}
b) / (a^{2}
b^{3})

(21 x^{5}) / (3 x^{4})

(6 x^{4})(4 y^{2}) / [ (3 x^{2})(16 y) ]

(4x – 12) / 4

(5x – 10) / (x + 2)

(x^{2}
– 4x – 12) / (x^{2}
– 2 x – 24)
Solution

Use rules of exponential to simplify the numerator first.
(a b^{2})(a^{3}
b) / (a^{2}
b^{3}) = (a^{4}
b^{3}) / (a^{2}
b^{3})
Rewrite as follows.
(a^{4}
/ a^{2}) (b^{3}
/ b^{3})
Use rule of quotient of exponentials to simplify.
= a^{2}

Rewrite as follows.
(21 x^{5}) / (3 x^{4}) = (21 / 3)(x^{5}
/ x^{4})
Simplify.
= 7 x

(6 x^{4})(4 y^{2}) / [ (3 x^{2})(16 y) ]
Multiply terms in numerator and denominator and simplify.
(6 x^{4})(4 y^{2}) / [ (3 x^{2})(16 y) ] = (24 x^{4}
y^{2}) / (48 x^{2}
y)
Rewrite as follows.
= (24 / 48)(x^{4}
/ x^{2})(y^{2}
/ y)
Simplify.
= (1 / 2) x^{2}
y

Factor 4 out in the numerator.
(4x – 12) / 4 = 4(x – 3) / 4
Simplify.
= x – 3

Factor 5 out in the numerator.
(5x – 10) / (x + 2) = – 5 (x + 2) / (x + 2)
Simplify.
= – 5

Factor numerator and denominator as follows.
(x^{2}
– 4x – 12) / (x^{2}
– 2x – 24) = [(x – 6)(x + 2)] / [(x – 6)(x + 4)]
Simplify.
= (x + 2) / (x + 4) , for all x not equal to 6

Solve for x the following linear equations.

2x = 6

6x – 8 = 4x + 4

4(x – 2) = 2(x + 3) + 7

0.1 x – 1.6 = 0.2 x + 2.3

– x / 5 = 2

(x – 4) / ( 6) = 3

(3x + 1) / (x – 2) = 3

x / 5 + (x – 1) / 3 = 1/5
Solution

Divide both sides of the equation by 2 and simplify.
2x / 2 = 6 / 2
x = 3

Add 8 to both sides and group like terms.
6x – 8 + 8 = 4x + 4 + 8
6x = 4x + 12
Add – 4x to both sides and group like terms.
6x – 4x = 4x + 12 – 4x
2x = 12
Divide both sides by 2 and simplify.
x = 6

Expand brackets.
4x – 8 = 2x + 6 + 7
Add 8 to both sides and group like terms.
4x – 8 + 8 = 2x + 6 + 7 + 8
4x = 2x + 21
Add – 2x to both sides and group like terms.
4x – 2x = 2x + 21 – 2x
2x = 21
Divide both sides by 2.
x = 21 / 2

Add 1.6 to both sides and simplify.
0.1 x – 1.6 = 0.2 x + 2.3
0.1 x – 1.6 + 1.6 = 0.2 x + 2.3 + 1.6
0.1 x = 0.2 x + 3.9
Add – 0.2 x to both sides and simplify.
0.1 x – 0.2 x = 0.2 x + 3.9 – 0.2 x
– 0.1 x = 3.9
Divide both sides by – 0.1 and simplify.
x = – 39

Multiply both sides by – 5 and simplify.
– 5( x / 5) = – 5(2)
x = – 10

Multiply both sides by – 6 and simplify.
(6)(x – 4) / ( 6) = (6)3
x – 4 = – 18
Add 4 to both sides and simplify.
x = – 14

Multiply both sides by (x – 2) and simplify.
(x – 2)(3x + 1) / (x – 2) = 3(x – 2)
Expand right term.
3x + 1 = 3x + 6
Add 3x to both sides and simplify.
– 3x + 1 + 3x = – 3x + 6 + 3x
1 = 6
The last statement is false and the equation has no solutions.

Multiply all terms by the LCM of 5 and 3 which is 15.
15(x / 5) + 15(x – 1) / 3 = 15(1 / 5)
Simplify and expand.
3x + 15x – 15 = 3
Group like terms and solve.
18 x = 3 + 15
18 x = 18
x = 1

Find any real solutions for the following quadratic equations.

2 x^{2}
– 8 = 0

x^{2}
= 5

2x^{2}
+ 5x – 7 = 0

(x – 2)(x + 3) = 0

(x + 7)(x – 1) = 9

x(x – 6) = 9
Solution

Divide all terms by 2.
2 x^{2}
/ 2 – 8 / 2 = 0 / 2
and simplify
x^{2}
– 4 = 0
Factor the right side.
(x – 2)(x + 2) = 0
Solve for x.
x – 2 = 0 or x = 2
x + 2 = 0 or x = 2
Solution set {2 , 2}

The given equation x^{2}
= 5 has no real solution since the square of real numbers is never negative.

Factor the left side as follows.
2x^{2}
+ 5x – 7 = 0
Factor
(2x + 7)(x – 1) = 0
Solve for x.
2x + 7 = 0 or x – 1 = 0
x = – 7/2 , x = 1, solution set:{7/2 , 1}

Solve for x.
(x – 2)(x + 3) = 0
x – 2 = 0 or x + 3 = 0
solution set: {3 , 2}

Expand left side.
x^{2}
+ 6x – 7 = 9
Rewrite the above equation with right side equal to 0.
x^{2}
+ 6x – 16 = 0
Factor left side.
(x + 8)(x – 2) = 0
Solve for x.
x + 8 = 0 or x – 2 = 0
solution set: {8 , 2}

Expand left side and rewrite with right side equal to zero.
x^{2}
– 6x + 9 = 0
Factor left side.
(x – 3)^{2}
= 0
Solve for x.
x – 3 = 0
solution set: {3}

Find any real solutions for the following equations.

x^{3}
– 1728 = 0

x^{3}
= – 64

√x = 1

√x = 5

√(x/100) = 4

√(200/x) = 2
Solution

Rewrite equation as.
x^{3}
= 1728
Take the cube root of each side.
(x^{3})^{1/3}
= (1728)^{1/3}
Simplify.
x = (1728)^{1/3}
= 12

Take the cube root of each side.
(x^{3})^{1/3}
= ( 64)^{1/3}
Simplify.
x = – 4

The equation √x= – 1 has no real solution because the square of a real number is greater than or equal to zero.

Square both sides.
(√x)^{2}
= 5^{2}
Simplify.
x = 25

Square both sides.
(√(x/100))^{2}
= 4^{2}
Simplify.
x / 100 = 16
Multiply both sides by 100 and simplify.
x = 1,600

Square both sides.
(√(200/x))^{2}
= 2^{2}
Simplify.
200 / x = 4
Multiply both sides by x and simplify.
x(200 / x) = 4 x
200 = 4 x
Solve for x.
x = 50

Evaluate for the given values of
a
and
b.

a^{2}
+ b^{2}
, for a = 2 and b = 2
2a – 3b , for a = 3 and b = 5

3a^{3}
– 4b^{4}
, for a = 1 and b = 2
Solution

Substitute a and b by their values and evaluate.
for a = 2 and b = 2
a^{2}
+ b^{2}
= 2^{2}
+ 2^{2}
= 8

Set a = – 3 and b = 5 in the given expression and evaluate.
 2a – 3b  =  2( 3) – 3(5)  =  6 – 15  =  21  = 21

Set a = – 1 and b = 2 in the given expression and evaluate.
3a^{3}
– 4b^{4}
= 3(1)^{3}
– 4(2)^{4}
= 3(1) – 4(16) = – 3 – 64 = – 67

Solve the following inequalities.

x + 3 < 0

x + 1 > x + 5

2(x – 2) < (x + 7)
Solution

Add 3 to both sides of the inequality and simplify.
x + 3 – 3 < 0 – 3
x < 3

Add x to both sides of the inequality and simplify.
x + 1 + x > – x + 5 + x
2x + 1 > 5
Add 1 to both sides of the inequality and simplify.
2x + 1 – 1 > 5 – 1
2x > 4
Divide both sides by 2.
x > 2

Expand brackets and group like terms.
2x – 4 < – x – 7
Add 4 to both sides and simplify.
2x – 4 + 4 < – x – 7 + 4
2x < – x – 3
Add x to both sides and simplify.
2x + x < – x – 3 + x
3x < – 3
Divide both sides by 3 and simplify.
x < – 1

For what value of the constant
k
does the quadratic equation
x^{2}
+2x = – 2k
have two distinct real solutions?
Solution
We first find write the given equation with right side equal to zero.
x^{2}
+2x + 2k = 0
We now calculate the discriminant D of the quadratic equation.
D = b^{2}
– 4 a c = 2^{2}
– 4 (1)(2k) = 4 – 8 k
For the solution to have two distinct real solution, D has to be positive. Hence
4 – 8 k > 0
Solve the inequality to get
k < 1/2

For what value of the constant
b
does the linear equation 2 x + b y = 2 have a slope equal to 2?
Solution
Solve for y and identify the slope
b y = – 2 x + 2
y = ( 2 / b) x + 2 / b
slope = ( 2 / b) = 2
Solve the equation ( 2 / b) = 2 for b
( 2 / b) = 2
2 = 2 b
b = – 1

What is the y intercept of the line
– 4 x + 6 y = – 12?
Solution
Set x = 0 in the equation and solve for y.
– 4 (0) + 6 y = – 12
6 y = – 12
y = – 2
y intercept: (0 , – 2)

What is the x intercept of the line
– 3 x + y = 3?
Solution
Set y = 0 in the equation and solve for x.
– 3 x + 0 = 3
x = 1
x intercept: (1 , 0)

What is point of intersection of the lines
x – y = 3
and
– 5 x – 2 y = – 22?
Solution
A point of intersection of two lines is solution to the equations of both lines. To find the point of intersection of the two lines, we need to solve the system of equations x – y = 3 and 5 x – 2 y = 22 simultaneously. Equation x – y = 3 can be solved for x to give
x = 3 + y
Substitute x by 3 + y in the equation – 5 x – 2 y = 22 and solve for y
5 (3 + y) – 2 y = – 22
15 – 5 y – 2 y = – 22
7 y = – 22 + 15
7 y = – 7
y = 1
Substitute x by 3 + y in the equation 5 x – 2 y = – 22 and solve for y
x = 3 + y = 3 + 1 = 4
Point of intersection: (4 , 1)

For what value of the constant
k
does the line
– 4 x + k y = 2
pass through the point
(2,3)?
Solution
For the line to pass through the point
(2,3), the ordered pair
(2,3)
must be a solution to the equation of the line. We substitute x by 2 abd y by – 3 in the equation.
– 4(2) + k(3) = 2
Solve the for k to obtain
k = – 10 / 3

What is the slope of the line with equation
y – 4 = 10?
Solution
Write the given equation in slope intercept form y = m x + b and identify the slope m.
y = 14
It is a horizontal line and therefore the slope is equal to 0.

What is the slope of the line with equation
2 x = 8?
Solution
The above equation may be written as
x = – 4
It is a vertical line and therefore the slope is undefined.

Find the x and y intercepts of the line with equation
x = – 3?
Solution
The above is a vertical line with x intercept only given by
(3 , 0)

Find the x and y intercepts of the line with equation
3 y – 6 = 3?
Solution
The given equation may be written as
y = 3
The above is a horizontal line with y intercept only given by
(0 , 3)

What is the slope of a line parallel to the x axis?
Solution
A line parallel to the x axis is a horizontal line and its slope is equal to 0.

What is the slope of a line perpendicular to the x axis?
Solution
A line perpendicular to the x axis is a vertical line and its slope is undefined.

More References and Links
Middle School Math (Grades 6, 7, 8, 9) – Free Questions and Problems With Answers
High School Math (Grades 10, 11 and 12) – Free Questions and Problems With Answers
Primary Math (Grades 4 and 5) with Free Questions and Problems With Answers
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