Solutions to Algebra Questions for Grade 9
Detailed solutions and full explanations to grade 9 algebra questions are presented.
-
Simplify the following algebraic expressions.
-
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.
\[
(x^{2}y)(xy^{2}) = (x^{2} \cdot x)(y \cdot y^{2})
\]
Using the rules of exponents:
\[
= x^{3} y^{3}
\]
-
Rewrite expression as follows.
\[
(-x^{2}y^{2})(xy^{2}) = -(x^{2} \cdot x)(y^{2} \cdot y^{2})
\]
Using the rules of exponents:
\[
= -x^{3} y^{4}
\]
-
Simplify the expressions.
-
Simplify numerator first using rules of exponents:
\[
(a b^{2})(a^{3} b) / (a^{2} b^{3}) = (a^{4} b^{3}) / (a^{2} b^{3})
\]
Rewrite as:
\[
\left(\frac{a^{4}}{a^{2}}\right) \left(\frac{b^{3}}{b^{3}}\right)
\]
Simplify:
\[
a^{2}
\]
-
Rewrite as:
\[
\frac{21x^{5}}{3x^{4}} = \left(\frac{21}{3}\right)\left(\frac{x^{5}}{x^{4}}\right)
\]
Simplify:
\[
7x
\]
-
Multiply and simplify:
\[
\frac{(6x^{4})(4y^{2})}{(3x^{2})(16y)} = \frac{24x^{4}y^{2}}{48x^{2}y}
\]
Rewrite as:
\[
\left(\frac{24}{48}\right)\left(\frac{x^{4}}{x^{2}}\right)\left(\frac{y^{2}}{y}\right)
\]
Simplify:
\[
\frac{1}{2}x^{2}y
\]
-
Factor 4 from numerator:
\[
\frac{4x - 12}{4} = \frac{4(x - 3)}{4} = x - 3
\]
-
Factor out \(-5\) from numerator:
\[
\frac{-5x - 10}{x + 2} = \frac{-5(x + 2)}{x + 2}
\]
Cancel \(x+2\) (assuming \(x \neq -2\)):
\[
-5
\]
-
Factor both numerator and denominator:
\[
\frac{x^{2} - 4x - 12}{x^{2} - 2x - 24}
\]
Numerator:
\[
x^{2} - 4x - 12 = (x - 6)(x + 2)
\]
Denominator:
\[
x^{2} - 2x - 24 = (x - 6)(x + 4)
\]
Simplify:
\[
\frac{(x - 6)(x + 2)}{(x - 6)(x + 4)} = \frac{x + 2}{x + 4} \quad (\text{for } x \neq 6, -4)
\]
-
Solve for \(x\) the following linear equations.
Solution
-
Divide both sides by 2:
\[
\frac{2x}{2} = \frac{6}{2}
\]
\[
x = 3
\]
-
Add 8 to both sides:
\[
6x - 8 + 8 = 4x + 4 + 8 \quad \Rightarrow \quad 6x = 4x + 12
\]
Subtract \(4x\) from both sides:
\[
6x - 4x = 12 \quad \Rightarrow \quad 2x = 12
\]
Divide by 2:
\[
x = 6
\]
-
Expand brackets:
\[
4x - 8 = 2x + 6 + 7
\]
Add 8 to both sides:
\[
4x = 2x + 21
\]
Subtract \(2x\):
\[
2x = 21
\]
Divide by 2:
\[
x = \frac{21}{2}
\]
-
Add 1.6 to both sides:
\[
0.1x = 0.2x + 3.9
\]
Subtract \(0.2x\):
\[
-0.1x = 3.9
\]
Divide by \(-0.1\):
\[
x = -39
\]
-
Multiply both sides by \(-5\):
\[
x = -10
\]
-
Multiply both sides by \(-6\):
\[
x - 4 = -18
\]
Add 4 to both sides:
\[
x = -14
\]
-
Multiply both sides by \(x - 2\):
\[
-3x + 1 = -3(x - 2) = -3x + 6
\]
Add \(3x\) to both sides:
\[
1 = 6
\]
This is false, so the equation has **no solution**.
-
Multiply through by the LCM of 5 and 3, which is 15:
\[
15\left(\frac{x}{5}\right) + 15\left(\frac{x - 1}{3}\right) = 15\left(\frac{1}{5}\right)
\]
Simplify:
\[
3x + 5(x - 1) = 3
\]
Expand:
\[
3x + 5x - 5 = 3
\]
Combine like terms:
\[
8x - 5 = 3
\]
Add 5:
\[
8x = 8
\]
Divide by 8:
\[
x = 1
\]
-
Find any real solutions for the following quadratic equations.
Solution
-
Divide all terms by 2:
\[
\frac{2x^2}{2} - \frac{8}{2} = \frac{0}{2}
\]
Simplify:
\[
x^2 - 4 = 0
\]
Factor:
\[
(x - 2)(x + 2) = 0
\]
Solve:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\]
Solution set: \(\{-2, 2\}\)
-
The equation
\[
x^2 = -5
\]
has **no real solution** since the square of a real number is never negative.
-
Factor:
\[
2x^2 + 5x - 7 = 0
\]
\[
(2x + 7)(x - 1) = 0
\]
Solve:
\[
2x + 7 = 0 \quad \Rightarrow \quad x = -\frac{7}{2}
\]
\[
x - 1 = 0 \quad \Rightarrow \quad x = 1
\]
Solution set: \(\left\{-\frac{7}{2}, 1\right\}\)
-
\[
(x - 2)(x + 3) = 0
\]
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
Solution set: \(\{-3, 2\}\)
-
Expand:
\[
x^2 + 6x - 7 = 9
\]
Bring all terms to one side:
\[
x^2 + 6x - 16 = 0
\]
Factor:
\[
(x + 8)(x - 2) = 0
\]
Solve:
\[
x + 8 = 0 \quad \Rightarrow \quad x = -8
\]
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
Solution set: \(\{-8, 2\}\)
-
Expand and bring terms to one side:
\[
x^2 - 6x + 9 = 0
\]
Factor:
\[
(x - 3)^2 = 0
\]
Solve:
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
Solution set: \(\{3\}\)
-
Find any real solutions for the following equations.
Solution
-
Rewrite:
\[
x^3 = 1728
\]
Take the cube root:
\[
(x^3)^{\frac{1}{3}} = (1728)^{\frac{1}{3}}
\]
Simplify:
\[
x = 12
\]
-
Take the cube root:
\[
(x^3)^{\frac{1}{3}} = (-64)^{\frac{1}{3}}
\]
Simplify:
\[
x = -4
\]
-
The equation
\[
\sqrt{x} = -1
\]
has **no real solution** because the square root of a real number is never negative.
-
Square both sides:
\[
(\sqrt{x})^2 = 5^2
\]
Simplify:
\[
x = 25
\]
-
Square both sides:
\[
\left(\sqrt{\frac{x}{100}}\right)^2 = 4^2
\]
Simplify:
\[
\frac{x}{100} = 16
\]
Multiply by 100:
\[
x = 1600
\]
-
Square both sides:
\[
\left(\sqrt{\frac{200}{x}}\right)^2 = 2^2
\]
Simplify:
\[
\frac{200}{x} = 4
\]
Multiply both sides by \(x\):
\[
200 = 4x
\]
Solve:
\[
x = 50
\]
-
Evaluate for the given values of a and b.
Solution
-
For \(a = 2\) and \(b = 2\):
\[
a^2 + b^2 = 2^2 + 2^2 = 4 + 4 = 8
\]
For \(a = -3\) and \(b = 5\):
\[
|2a - 3b| = |2(-3) - 3(5)| = |-6 - 15| = |-21| = 21
\]
-
For \(a = -1\) and \(b = -2\):
\[
3a^3 - 4b^4 = 3(-1)^3 - 4(-2)^4
\]
\[
= 3(-1) - 4(16) = -3 - 64 = -67
\]
-
Solve the following inequalities.
Solution
-
\[
x + 3 \lt 0
\]
Subtract 3 from both sides:
\[
x \lt -3
\]
-
\[
x + 1 > -x + 5
\]
Add \(x\) to both sides:
\[
2x + 1 > 5
\]
Subtract 1 from both sides:
\[
2x > 4
\]
Divide by 2:
\[
x > 2
\]
-
\[
2(x - 2) \lt -(x + 7)
\]
Expand:
\[
2x - 4 \lt -x - 7
\]
Add 4 to both sides:
\[
2x \lt -x - 3
\]
Add \(x\) to both sides:
\[
3x \lt -3
\]
Divide by 3:
\[
x \lt -1
\]
-
For what value of the constant k does the quadratic equation \(x^2 + 2x = -2k\) have two distinct real solutions?
Solution
Rewrite with the right side equal to zero:
\[
x^2 + 2x + 2k = 0
\]
The discriminant is:
\[
D = b^2 - 4ac = 2^2 - 4(1)(2k) = 4 - 8k
\]
For two distinct real solutions, \(D > 0\):
\[
4 - 8k > 0
\]
\[
k \lt \frac{1}{2}
\]
-
For what value of the constant b does the linear equation \(2x + by = 2\) have a slope equal to 2?
Solution
Solve for \(y\):
\[
by = -2x + 2
\]
\[
y = \frac{-2}{b}x + \frac{2}{b}
\]
The slope is:
\[
\frac{-2}{b} = 2
\]
Multiply through by \(b\):
\[
-2 = 2b
\]
\[
b = -1
\]
-
What is the y-intercept of the line \(-4x + 6y = -12\)?
Solution
Set \(x = 0\):
\[
-4(0) + 6y = -12
\]
\[
6y = -12
\]
\[
y = -2
\]
y-intercept: \((0, -2)\)
-
What is the x-intercept of the line \(-3x + y = 3\)?
Solution
Set \(y = 0\):
\[
-3x + 0 = 3
\]
\[
-3x = 3
\]
\[
x = -1
\]
x-intercept: \((-1, 0)\)
-
What is the point of intersection of the lines \(x - y = 3\) and \(-5x - 2y = -22\)?
Solution
From \(x - y = 3\):
\[
x = 3 + y
\]
Substitute into \(-5x - 2y = -22\):
\[
-5(3 + y) - 2y = -22
\]
\[
-15 - 5y - 2y = -22
\]
\[
-7y = -22 + 15
\]
\[
-7y = -7
\]
\[
y = 1
\]
Substitute into \(x = 3 + y\):
\[
x = 3 + 1 = 4
\]
Point of intersection: \((4, 1)\)
-
For what value of the constant \(k\) does the line \(-4x + ky = 2\) pass through the point \((2, -3)\)?
Solution
Substitute \(x = 2\) and \(y = -3\) into the equation:
\[
-4(2) + k(-3) = 2
\]
\[
-8 - 3k = 2
\]
\[
-3k = 10
\]
\[
k = -\frac{10}{3}
\]
-
What is the slope of the line \(y - 4 = 10\)?
Solution
Rewrite in slope-intercept form:
\[
y = 14
\]
This is a horizontal line, so:
\[
\text{slope} = 0
\]
-
What is the slope of the line \(2x = -8\)?
Solution
Rewrite:
\[
x = -4
\]
This is a vertical line, so the slope is undefined.
-
Find the x- and y-intercepts of the line \(x = -3\).
Solution
This is a vertical line with only an x-intercept:
\[
(-3, 0)
\]
-
Find the x- and y-intercepts of the line \(3y - 6 = 3\).
Solution
Rewrite:
\[
3y - 6 = 3
\]
\[
3y = 9
\]
\[
y = 3
\]
This is a horizontal line with only a y-intercept:
\[
(0, 3)
\]
-
What is the slope of a line parallel to the x-axis?
Solution
A line parallel to the x-axis is horizontal:
\[
\text{slope} = 0
\]
-
What is the slope of a line perpendicular to the x-axis?
Solution
A line perpendicular to the x-axis is vertical, so the slope is undefined.
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