2/3 X 5/6 X 14

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.
1. - 6x + 5 + 12x -6
2. ii(x - 9) + six(-x + 2) + 4x
3. 3x^two + 12 + 9x - 20 + 6x² - x
4. (ten + two)(x + iv) + (x + 5)(-x - i)
5. 1.2(x - 9) - 2.iii(ten + 4)
6. (x²y)(xy^two)
7. (-x²y^two)(xy²)
Solution
1. Group like terms and simplify.
  - 6x + five + 12x -6 = (- 6x + 12x) + (5 - 6)
  = 6x - 1
2. Expand brackets.
  ii(x - 9) + 6(-x + 2) + 4x = 2x - xviii - 6x + 12 + 4x
  Group like terms and simplify.
  = (2x - 6x + 4x) + (- 18 + 12) = - 6
3. Group like terms and simplify.
  3x^two + 12 + 9x - 20 + 6x² - x
  = (3x² + 6x²) + (9x - x) + (12 - 20)
  = 9xⁱⁱ + 8x - 8
4. Expand brackets.
  (ten + two)(x + four) + (x + v)(- 10 - one)
  = x² + 4x + 2x + 8 - 10² - x - 5x - 5
  Group like terms.
  = (x² - xⁱⁱ) + (4x + 2x - x - 5x) + (8 - 5)
  = iii
5. Aggrandize and group.
  1.2(x - ix) - 2.3(x + 4)
  = 1.2x - ten.8 - 2.3x - 9.two
  = -1.1x - twenty
6. Rewrite every bit follows.
  (10²y)(xyⁱⁱ) = (x² x)(y yⁱⁱ)
  Use rules of exponential.
  = x³ y³
7. Rewrite expression equally follows.
  (-x²y^two)(xy²) = -(x² x)( y^two yⁱⁱ)
  Use rules of exponential.
  = - x³ y⁴
Simplify the expressions.
1. (a b^two)(a³ b) / (a² b³)
2. (21 x^v) / (3 10^iv)
3. (6 x⁴)(4 y²) / [ (three x²)(xvi y) ]
4. (4x - 12) / 4
5. (-5x - 10) / (10 + 2)
6. (10^two - 4x - 12) / (tenⁱⁱ - 2 x - 24)
Solution
1. Use rules of exponential to simplify the numerator first.
  (a b²)(a³ b) / (a^two b³) = (a⁴ b^three) / (a² b³)
  Rewrite every bit follows.
  (a⁴ / a²) (b³ / b³)
  Employ rule of quotient of exponentials to simplify.
  = a²
2. Rewrite equally follows.
  (21 ten^v) / (three x⁴) = (21 / iii)(x^v / x⁴)
  Simplify.
  = 7 ten
3. (6 x^iv)(4 y^two) / [ (3 x²)(sixteen y) ]
  Multiply terms in numerator and denominator and simplify.
  (half-dozen x⁴)(iv y²) / [ (3 x²)(16 y) ] = (24 x^iv y²) / (48 10² y)
  Rewrite as follows.
  = (24 / 48)(10^four / x²)(y² / y)
  Simplify.
  = (1 / two) x² y
4. Factor 4 out in the numerator.
  (4x - 12) / 4 = iv(x - three) / 4
  Simplify.
  = x - three
5. Factor -five out in the numerator.
  (-5x - ten) / (x + ii) = - 5 (x + 2) / (ten + two)
  Simplify.
  = - v
6. Gene numerator and denominator as follows.
  (x^two - 4x - 12) / (x^two - 2x - 24) = [(ten - 6)(x + two)] / [(ten - vi)(x + four)]
  Simplify.
  = (x + 2) / (x + iv) , for all x not equal to half dozen
Solve for ten the following linear equations.
1. 2x = half-dozen
2. 6x - 8 = 4x + 4
3. 4(x - 2) = ii(x + 3) + 7
4. 0.i x - 1.6 = 0.two x + 2.3
5. - ten / 5 = 2
6. (x - iv) / (- six) = 3
7. (-3x + 1) / (10 - 2) = -3
8. x / five + (x - 1) / 3 = 1/5
Solution
1. Separate both sides of the equation by two and simplify.
  2x / two = 6 / 2
  ten = three
2. Add 8 to both sides and group like terms.
  6x - 8 + 8 = 4x + iv + 8
  6x = 4x + 12
  Add - 4x to both sides and group like terms.
  6x - 4x = 4x + 12 - 4x
  2x = 12
  Dissever both sides by two and simplify.
  x = half-dozen
3. Aggrandize brackets.
  4x - 8 = 2x + 6 + seven
  Add viii to both sides and group like terms.
  4x - 8 + 8 = 2x + six + 7 + 8
  4x = 2x + 21
  Add together - 2x to both sides and group like terms.
  4x - 2x = 2x + 21 - 2x
  2x = 21
  Split up both sides by 2.
  ten = 21 / 2
4. Add together 1.6 to both sides and simplify.
  0.1 x - one.6 = 0.2 10 + 2.iii
  0.1 x - 1.6 + 1.6 = 0.2 x + ii.iii + 1.6
  0.ane x = 0.2 x + 3.ix
  Add - 0.ii x to both sides and simplify.
  0.one ten - 0.2 x = 0.2 x + iii.9 - 0.ii 10
  - 0.one x = 3.9
  Divide both sides past - 0.1 and simplify.
  x = - 39
5. Multiply both sides past - 5 and simplify.
  - v(- x / 5) = - 5(2)
  x = - 10
6. Multiply both sides by - six and simplify.
  (-6)(x - 4) / (- 6) = (-6)3
  x - 4 = - eighteen
  Add together 4 to both sides and simplify.
  ten = - 14
7. Multiply both sides by (x - ii) and simplify.
  (x - 2)(-3x + 1) / (x - 2) = -3(10 - 2)
  Expand correct term.
  -3x + 1 = -3x + 6
  Add 3x to both sides and simplify.
  - 3x + one + 3x = - 3x + half-dozen + 3x
  1 = 6
  The last statement is faux and the equation has no solutions.
8. Multiply all terms by the LCM of 5 and 3 which is 15.
  15(10 / 5) + 15(x - one) / iii = 15(i / 5)
  Simplify and aggrandize.
  3x + 15x - fifteen = three
  Grouping like terms and solve.
  18 x = 3 + 15
  xviii ten = 18
  x = 1
Find any existent solutions for the following quadratic equations.
1. 2 x² - 8 = 0
2. x² = -v
3. 2x^two + 5x - 7 = 0
4. (x - two)(x + 3) = 0
5. (x + vii)(10 - 1) = 9
6. 10(x - 6) = -9
Solution
1. Divide all terms by 2.
  2 10ⁱⁱ / 2 - eight / 2 = 0 / 2
  and simplify
  x² - iv = 0
  Gene the correct side.
  (x - two)(x + two) = 0
  Solve for x.
  x - 2 = 0 or ten = 2
  x + 2 = 0 or x = -2
  Solution set {-2 , two}
2. The given equation 10² = -5 has no real solution since the square of real numbers is never negative.
3. Factor the left side equally follows.
  2x² + 5x - vii = 0
  Cistron
  (2x + 7)(x - 1) = 0
  Solve for x.
  2x + seven = 0 or 10 - one = 0
  ten = - seven/2 , x = ane, solution prepare:{-seven/2 , 1}
4. Solve for x.
  (x - 2)(x + three) = 0
  ten - two = 0 or x + 3 = 0
  solution set up: {-3 , 2}
5. Aggrandize left side.
  x² + 6x - 7 = 9
  Rewrite the in a higher place equation with right side equal to 0.
  x² + 6x - 16 = 0
  Factor left side.
  (x + eight)(x - two) = 0
  Solve for x.
  x + 8 = 0 or x - two = 0
  solution set: {-8 , 2}
6. Aggrandize left side and rewrite with correct side equal to zero.
  x² - 6x + 9 = 0
  Factor left side.
  (x - 3)ⁱⁱ = 0
  Solve for 10.
  x - 3 = 0
  solution set: {three}
Find whatever real solutions for the following equations.
1. x³ - 1728 = 0
2. x³ = - 64
3. √x = -1
4. √x = 5
5. √(10/100) = 4
6. √(200/x) = ii
Solution
1. Rewrite equation as.
  10³ = 1728
  Take the cube root of each side.
  (x^three)^1/3 = (1728)^i/3
  Simplify.
  x = (1728)^ane/3 = 12
2. Take the cube root of each side.
  (x³)^1/iii = (- 64)^1/3
  Simplify.
  x = - 4
3. The equation √10= - 1 has no real solution because the foursquare of a real number is greater than or equal to zero.
4. Square both sides.
  (√x)^two = 5²
  Simplify.
  x = 25
5. Square both sides.
  (√(x/100))² = 4²
  Simplify.
  ten / 100 = sixteen
  Multiply both sides by 100 and simplify.
  x = 1,600
6. Square both sides.
  (√(200/x))² = 2ⁱⁱ
  Simplify.
  200 / x = iv
  Multiply both sides past ten and simplify.
  10(200 / x) = iv x
  200 = 4 x
  Solve for x.
  x = 50
Evaluate for the given values of a and b.
1. a² + bⁱⁱ , for a = 2 and b = 2
  |2a - 3b| , for a = -3 and b = 5
2. 3aⁱⁱⁱ - 4b^four , for a = -1 and b = -2
Solution
1. Substitute a and b by their values and evaluate.
  for a = 2 and b = 2
  a² + b² = 2ⁱⁱ + ii² = 8
2. Prepare a = - three and b = 5 in the given expression and evaluate.
  | 2a - 3b | = | 2( -three) - three(five) | = | -6 - 15 | = | -21 | = 21
3. Set a = - 1 and b = -two in the given expression and evaluate.
  3a³ - 4b⁴ = 3(-1)³ - 4(-ii)^iv = 3(-1) - 4(xvi) = - 3 - 64 = - 67
Solve the following inequalities.
1. x + iii < 0
2. x + 1 > -10 + five
3. 2(x - ii) < -(ten + 7)
Solution
1. Add -3 to both sides of the inequality and simplify.
  x + iii - 3 < 0 - three
  x < -3
2. Add x to both sides of the inequality and simplify.
  x + i + x > - x + v + 10
  2x + i > five
  Add -1 to both sides of the inequality and simplify.
  2x + 1 - i > 5 - ane
  2x > 4
  Divide both sides by 2.
  x > two
3. Expand brackets and group like terms.
  2x - four < - x - 7
  Add 4 to both sides and simplify.
  2x - 4 + four < - ten - 7 + 4
  2x < - x - 3
  Add x to both sides and simplify.
  2x + x < - 10 - 3 + x
  3x < - iii
  Divide both sides past three and simplify.
  x < - ane
For what value of the constant k does the quadratic equation x² +2x = - 2k have ii singled-out real solutions?
Solution
Nosotros first discover write the given equation with right side equal to aught.
10^two +2x + 2k = 0
We now calculate the discriminant D of the quadratic equation.
D = b² - 4 a c = two² - four (1)(2k) = four - eight k
For the solution to have two singled-out real solution, D has to be positive. Hence
4 - 8 m > 0
Solve the inequality to get
k < one/2
For what value of the constant b does the linear equation 2 x + b y = two have a slope equal to two?
Solution
Solve for y and place the gradient
b y = - 2 x + 2
y = (- ii / b) x + 2 / b
slope = (- 2 / b) = 2
Solve the equation (- 2 / b) = two for b
(- two / b) = two
-2 = 2 b
b = - one
What is the y intercept of the line - four x + 6 y = - 12?
Solution
Ready 10 = 0 in the equation and solve for y.
- 4 (0) + vi y = - 12
vi y = - 12
y = - 2
y intercept: (0 , - ii)
What is the ten intercept of the line - three x + y = 3?
Solution
Fix y = 0 in the equation and solve for x.
- 3 x + 0 = 3
ten = -1
10 intercept: (-1 , 0)
What is point of intersection of the lines ten - y = three 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 bespeak of intersection of the ii lines, nosotros need to solve the organization of equations x - y = iii and -v x - 2 y = -22 simultaneously. Equation ten - y = 3 can be solved for x to give
10 = 3 + y
Substitute x past 3 + y in the equation - 5 x - ii y = -22 and solve for y
-5 (3 + y) - two y = - 22
-fifteen - 5 y - 2 y = - 22
-7 y = - 22 + 15
-vii y = - vii
y = 1
Substitute x past 3 + y in the equation -5 ten - 2 y = - 22 and solve for y
x = 3 + y = 3 + 1 = 4
Signal of intersection: (four , 1)
For what value of the abiding grand does the line - 4 x + chiliad y = 2 pass through the point (2,-three)?
Solution
For the line to laissez passer through the point (two,-3), the ordered pair (2,-3) must exist a solution to the equation of the line. We substitute ten by 2 abd y by - 3 in the equation.
- 4(2) + 1000(-3) = 2
Solve the for k to obtain
k = - x / 3
What is the slope of the line with equation y - 4 = 10?
Solution
Write the given equation in slope intercept form y = yard x + b and identify the gradient m.
y = xiv
Information technology 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
10 = - 4
It is a vertical line and therefore the gradient is undefined.
Find the x and y intercepts of the line with equation 10 = - 3?
Solution
The higher up is a vertical line with x intercept only given past
(-three , 0)
Notice the x and y intercepts of the line with equation three y - half dozen = 3?
Solution
The given equation may exist written equally
y = 3
The above is a horizontal line with y intercept merely given by
(0 , three)
What is the slope of a line parallel to the x axis?
Solution
A line parallel to the 10 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.