Algebra Questions with Solutions for Grade 9

A. Group like terms and simplify: \[ -6x + 5 + 12x - 6 = (-6x + 12x) + (5 - 6) = 6x - 1 \]

B. Expand brackets: \[ 2(x - 9) + 6(-x + 2) + 4x = 2x - 18 - 6x + 12 + 4x \] Group like terms and simplify: \[ (2x - 6x + 4x) + (-18 + 12) = -6 \]

C. Group like terms and simplify: \[ 3x^2 + 12 + 9x - 20 + 6x^2 - x \] \[ = (3x^2 + 6x^2) + (9x - x) + (12 - 20) = 9x^2 + 8x - 8 \]

D. Expand brackets: \[ (x + 2)(x + 4) + (x + 5)(-x - 1) = x^2 + 4x + 2x + 8 - x^2 - x - 5x - 5 \] Group like terms: \[ (x^2 - x^2) + (4x + 2x - x - 5x) + (8 - 5) = 3 \]

E. Expand and group: \[ 1.2(x - 9) - 2.3(x + 4) = 1.2x - 10.8 - 2.3x - 9.2 = -1.1x - 20 \]

F. Rewrite using exponential rules: \[ (x^2y)(xy^2) = (x^2 \cdot x)(y \cdot y^2) = x^3y^3 \]

G. Rewrite expression using exponential rules: \[ (-x^2y^2)(xy^2) = -(x^2 \cdot x)(y^2 \cdot y^2) = -x^3y^4 \]

A. \( \dfrac{(a b^2)(a^3 b)}{a^2 b^3} \)

Simplify the numerator first: \[ \dfrac{a^4 b^3}{a^2 b^3} \] Rewrite as: \[ \dfrac{a^4}{a^2} \cdot \dfrac{b^3}{b^3} \] Simplify using the quotient rule for exponents: \[ a^2 \] B. \( \dfrac{21 x^5}{3 x^4} \)

Rewrite as: \[ \dfrac{21}{3} \cdot \dfrac{x^5}{x^4} \] Simplify: \[ 7x \] C. \( \dfrac{(6 x^4)(4 y^2)}{(3 x^2)(16 y)} \)

Multiply terms in the numerator and denominator: \[ \dfrac{24 x^4 y^2}{48 x^2 y} \] Rewrite as: \[ \dfrac{24}{48} \cdot \dfrac{x^4}{x^2} \cdot \dfrac{y^2}{y} \] Simplify: \[ \dfrac{1}{2} x^2 y \] D. \( \quad \dfrac{4x - 12}{4} \)

Factor 4 out of the numerator: \[ \dfrac{4(x - 3)}{4} \] Simplify: \[ x - 3 \] E. \( \dfrac{-5x - 10}{x + 2} \)

Factor -5 out of the numerator: \[ \dfrac{-5(x + 2)}{x + 2} \] Simplify: \[ -5 \]

F. \( \dfrac{x^2 - 4x - 12}{x^2 - 2x - 24} \)

Factor both the numerator and denominator: \[ \dfrac{(x - 6)(x + 2)}{(x - 6)(x + 4)} \] Simplify: \[ \dfrac{x + 2}{x + 4} \quad \text{(for all } x \neq 6 \text{)} \]

A. Divide both sides of the equation by 2 and simplify. \[ \frac{2x}{2} = \frac{6}{2} \] \[ x = 3 \] B. Add 8 to both sides and group like terms. \[ 6x - 8 + 8 = 4x + 4 + 8 \] Group like terms \[ 6x = 4x + 12 \] Add \( -4 x \) to both sides \[ 6x - 4x = 4x + 12 - 4x \] Group like terms \[ 2x = 12 \] Divide both sides by 2 and simplify. \[ x = 6 \] C. Expand brackets in \( 4(x - 2) = 2(x + 3) + 7 \). \[ 4x - 8 = 2x + 6 + 7 \] Add \(8\) to both sides and group like terms. \[ 4x - 8 + 8 = 2x + 6 + 7 + 8 \] Group like terms \[ 4x = 2x + 21 \] Add \( -2x \) to both sides and group like terms. \[ 4x - 2x = 2x + 21 - 2x \] Group like terms \[ 2x = 21 \] Divide both sides by 2. \[ x = \frac{21}{2} \] D. Add 1.6 to both sides and simplify. \[ 0.1x - 1.6 + 1.6 = 0.2x + 2.3 + 1.6 \] Group like terms \[ 0.1x = 0.2x + 3.9 \] Add \( -0.2x \) to both sides and simplify. \[ 0.1x - 0.2x = 0.2x + 3.9 - 0.2x \] Group like terms \[ -0.1 x = 3.9 \] Divide both sides by \( -0.1 \) and simplify. \[ x = -39 \] E. Multiply both sides by \( -5 \) and simplify. \[ -5 \left(\frac{-x}{5} \right) = -5(2) \] \[ x = -10 \]

F. Multiply both sides by \( -6 \) and simplify. \[ \frac{(-6)(x - 4)}{-6} = (-6)3 \] \[ x - 4 = -18 \] Add \( 4 \) to both sides. \[ x - 4 + 4 = -18 + 4 \] Group like terms \[ x = -14 \]

G. Multiply both sides by \( (x - 2) \). \[ \frac{(x - 2)(-3x + 1)}{x - 2} = -3(x - 2) \] Simplify the left side \[ -3x + 1 = -3(x - 2) \] Expand right term. \[ -3 x + 1 = -3 x + 6 \] Add \( 3x \) to both sides. \[ -3x + 1 + 3x = -3x + 6 + 3x \] Group like terms and simplify \[ 1 = 6 \] The last statement is false and the equation has no solutions

H. Multiply all terms 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 to eliminate the denominators \[ 3 x + 5(x - 1) = 3 \times 1 \] Expand. \[ 3x + 5x - 5 = 3 \] Group like terms and solve. \[ 8 x = 3 + 5 \] \[ 8x = 8 \] \[ x = 1 \]

A. Divide all terms by 2. \[ \frac{2x^2}{2} - \frac{8}{2} = \frac{0}{2} \] Simplify \[ \quad x^2 - 4 = 0 \] Factor the leftt side: \[ (x - 2)(x + 2) = 0 \] Solve for \( x \): \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] Solution set is: \[ \{-2, 2\} \] B. The given equation \( x^2 = -5 \) has no real solution, since the square of real numbers is never negative.

C. Factor the left side as follows: \[ 2x^2 + 5x - 7 = 0 \] Factor the left side. \[ (2x + 7)(x - 1) = 0 \] Solve for \( x \) by setting each of the factors equal to zero: \[ 2x + 7 = 0 \quad \Rightarrow \quad x = -\frac{7}{2} \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \] Solution set: \[ \left\{-\frac{7}{2}, 1\right\} \] D. The equation is in factored form and is solved by setting each of the factors equal to zero. \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] Solution set: \[ \{-3, 2\} \] E. Expand the left side: \[ x^2 + 6x - 7 = 9 \] Rewrite the equation with the right side equal to 0: \[ x^2 + 6x - 16 = 0 \] Factor the left side: \[ (x + 8)(x - 2) = 0 \] Solve for \( x \) by setting each of the factors equal to zero: \] \[ x + 8 = 0 \quad \Rightarrow \quad x = -8 \] \[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \] Solution set: \[ \{ -8, 2 \} \]

F. Expand the left side and rewrite with the right side equal to zero: \[ x^2 - 6x + 9 = 0 \] Factor the left side: \[ (x - 3)^2 = 0 \] Which gives \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] Solution set: \[ \{3 \} \]

A. Rewrite the equation as: \[ x^3 = 1728 \] Take the cube root of each side: \[ \sqrt[3]{x^3} = \sqrt[3] {1728} \] Simplify: \[ x = \sqrt[3] {1728} = 12 \] B. Take the cube root of each side: \[ \sqrt[3] {x^3} = \sqrt[3]{-64)} \] Simplify: \[ x = -4 \] C. The equation \( \sqrt{x} = -1 \) has no real solution because the square of a real number is greater than or equal to zero.

D. Square both sides: \[ (\sqrt{x})^2 = 5^2 \] Simplify: \[ x = 25 \] E. Square both sides: \[ \left(\sqrt{\frac{x}{100}}\right)^2 = 4^2 \] Simplify: \[ \frac{x}{100} = 16 \] Multiply both sides by 100 and simplify: \[ x = 1600 \]

F. Square both sides: \[ \left(\sqrt{\frac{200}{x}}\right)^2 = 2^2 \] Simplify: \[ \frac{200}{x} = 4 \] Multiply both sides by \( x \) and simplify: \[ x \cdot \frac{200}{x} = 4x \] \[ 200 = 4x \] Solve for \( x \): \[ x = 50 \]

A. Substitute \( a \) and \( b \) by their values and evaluate. For \( a = 2 \) and \( b = 2 \): \[ a^2 + b^2 = 2^2 + 2^2 = 4 + 4 = 8 \] B. Set \( a = -3 \) and \( b = 5 \) in the given expression and evaluate. \[ | 2a - 3b | = | 2(-3) - 3(5) | = | -6 - 15 | = | -21 | = 21 \] C. Set \( a = -1 \) and \( b = -2 \) in the given expression and evaluate. \[ 3a^3 - 4b^4 = 3(-1)^3 - 4(-2)^4 = 3(-1) - 4(16) = -3 - 64 = -67 \]

A. Add \( -3 \) to both sides of the inequality and simplify. \[ x + 3 - 3 \lt 0 - 3 \] Group like terms to obtain the solution set: \[ x \lt -3 \] B. Add \( x \) to both sides of the inequality \[ x + 1 + x \gt -x + 5 + x \] Group like terms \[ 2x + 1 \gt 5 \] Add \( -1 \) to both sides of the inequality and simplify. \[ 2x + 1 - 1 \gt 5 - 1 \] Group like terms \[ 2x \gt 4 \] Divide both sides by \( 2 \) to obtain the solution set: \[ x \gt 2 \] C. Expand brackets and group like terms. \[ 2x - 4 \lt -x - 7 \] Add \( 4 \) to both sides. \[ 2x - 4 + 4 \lt -x - 7 + 4 \] Group like terms \[ 2x \lt -x - 3 \] Add \( x \) to both sides. \[ 2x + x \lt -x - 3 + x \] Group like terms \[ 3x \lt -3 \] Divide both sides by \( 3 \) and simplify to obtain the solution set \[ x \lt -1 \]

We first write the given equation with the right side equal to zero: \[ x^2 + 2x + 2k = 0 \] Next, we calculate the discriminant \( D \) of the quadratic equation: \[ D = b^2 - 4ac = 2^2 - 4(1)(2k) = 4 - 8k \] For the equation to have two distinct real solutions, the discriminant must be positive: \[ 4 - 8k > 0 \] Solving the inequality: \[ k \lt \frac{1}{2} \]

Solve for \( y \) and identify the slope: \[ b y = -2 x + 2 \] \[ y = \frac{-2}{b} x + \frac{2}{b} \] \[ \text{slope} = \frac{-2}{b} = 2 \] Solve the equation \( \frac{-2}{b} = 2 \) for \( b \): \[ \frac{-2}{b} = 2 \] \[ -2 = 2b \] \[ b = -1 \]

Set \( x = 0 \) in the equation and solve for \( y \). \[ -4(0) + 6y = -12 \] \[ 6y = -12 \] \[ y = -2 \] y-intercept: \( (0, -2) \)

Set y = 0 in the equation and solve for x.
- 3 x + 0 = 3
x = -1
x intercept: (-1 , 0)

A point of intersection of two lines is the 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 \[ -5x - 2y = -22 \] First, solve the equation \(x - y = 3\) for \(x\): \[ x = 3 + y \] Next, substitute \(x = 3 + y\) into the equation \(-5x - 2y = -22\) and solve for \(y\): \[ -5(3 + y) - 2y = -22 \] Simplify the equation: \[ -15 - 5y - 2y = -22 \] Combine like terms: \[ -7y = -22 + 15 \] Simplify further: \[ -7y = -7 \] Solve for \(y\): \[ y = 1 \] Now, substitute \(y = 1\) into \(x = 3 + y\) to find \(x\): \[ x = 3 + 1 = 4 \] Thus, the point of intersection is \[ (4, 1) \]

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 \( 2 \) and \( y \) by \( -3 \) in the equation. \[ -4(2) + k(-3) = 2 \] Solve for \( k \) to obtain \[ k = \frac{-10}{3} \]

Write the given equation in slope-intercept form \( y = mx + b \) and identify the slope \( m \). \[ y = 14 \] It is a horizontal line, and therefore the slope is equal to zero.

Divide all terms of the given equation by 2 \[ \dfrac{2 x}{2} = \dfrac{- 8}{2} \] Simplify \[ x = -4 \] It is a vertical line and therefore the slope is undefined.

\( x = - 3\) is a vertical line with \( x \) intercept only given by \[ (-3 , 0) \] A vertical line has no y intercept.

The given equation may be written as \[ 3 y = 9 \] Divide all terms by \( 3 \) to obtain \[ y = 3 \] The above is a horizontal line with \( y \) intercept only given by \[ (0 , 3) \] A horizontal line has no \( x \) intercept.

A line parallel to the \(x\) axis is a horizontal line and its slope is equal to zero.

A line perpendicular to the \( x \) axis is a vertical line and its slope is undefined.