Solutions to Algebra Questions and Problems for Grade 6

Solutions

Like terms have the same variable with equal exponents. Also all numbers are like terms.

A) The terms \(6x\) and \(12x\) have same variable \(x\) with exponents equal to 1 and are therefore like terms.

The terms \(5\) and \(-6\) are numbers and therefore like terms.

B) The terms \(2x^2\) and \(9x^2\) have same variable \(x\) with exponents equal to 2, they are like terms.

The terms \(-4\) and \(+9\) are numbers and therefore like terms.

C) The terms \(\frac{x}{5}\) and \(\frac{x}{7}\) have the same variable with exponents equal to 1, they are like terms.

D) The terms \(0.2x\), \(1.2x\), and \(\frac{x}{2}\) have the same variable \(x\) with exponents equal to 1; they are like terms.

E) The terms \(5x\) and \(7x\) are like terms.
The terms \(-8\) and \(-4\) are like terms.
The terms \(-2x^2\) and \(+9x^2\) are like terms.

F) There are no like terms in this expression.

G) The terms \(5ab\) and \(6ba\) are like terms.
Substitute \(x\) by the given numerical value and simplify.

A) \(6(2) + 5 = 12 + 5 = 17\)

B) \(12(1)^2 + 5(1) - 2 = 12(1) + 5 - 2 = 12 + 5 - 2 = 15\)

C) \(2(0 + 7) + 0 = 2(7) = 14\)

D) \(2(2) + 3(4) - 7 = 4 + 12 - 7 = 9\)
Notes:
1. We may add or subtract like terms as follows: \[2x + 6x = (2 + 6)x\] by factoring first \[= 8x\] simplifying the terms in brackets.
2. We use distributive property to expand an expression as follows: \(a(b + c) = ab + ac\)
A) \(3x + 5x\)

\(= (3 + 5)x\), factor \(x\) out

\(= 8x\), simplify

B) \(2(x + 7) + x\)

\(= 2(x) + 2(7) + x = 2x + 14 + x\), expand and simplify

\(= (2x + x) + 14\), group like terms

\(= (2 + 1)x + 14 = 3x + 14\), simplify

C) \(2(x + 3) + 3(x + 5) + 3\)

\(= 2(x) + 2(3) + 3(x) + 3(5) + 3 = 2x + 6 + 3x + 15 + 3\), expand and simplify

\(= (2x + 3x) + (6 + 15 + 3)\), group like terms

\(= (2 + 3)x + 24 = 5x + 24\), simplify

D) \(2(a + 1) + 5b + 3(a + b) + 3\)

\(= 2(a) + 2(1) + 5b + 3(a) + 3(b) + 3 = 2a + 2 + 5b + 3a + 3b + 3\), expand and simplify

\(= (2a + 3a) + (5b + 3b) + (2 + 3)\), group like terms

\(= (2 + 3)a + (5 + 3)b + 5 = 5a + 8b + 5\), simplify
Notes:
1. To factor an expression is to write it as a product.
2. The distributive property may be used to expand an algebraic expression: \(a(x + y) = ax + ay\)
3. The same distributive property may be used as: \(\color{red}{a}x + \color{red}{a}y = \color{red}{a}(x + y)\) to factor an expression where \(\color{red}{a}\) is called the common factor
A) \(3x + 3\)

\(= 3(x) + 3(1)\), 3 is a common factor

\(= 3(x + 1)\), factored form

B) \(8x + 4\)

\(= 4(2)(x) + 4\), write 8 as 4(2)

\(= 4(2x) + 4(1)\), 4 is a common factor

\(= 4(2x + 1)\), factored form

C) \(ax + 3a\), a is a common factor

\(= a(x + 3)\), factored form

D) \((x + 1)y + 4(x + 1)\)

\(= (x + 1)(y) + (x + 1)(4)\), x + 1 is a common factor

\(= (x + 1)(y + 4)\), factored form

E) \(x + 2 + bx + 2b\)

\(= (x + 2) + b(x + 2)\), factor b out in bx + 2b

\(= (x + 2)(1) + (x + 2)b\), x + 2 is now a common factor

\(= (x + 2)(1 + b)\), factored form

In order to check factoring, expand the factored form, simplify and make sure it is equivalent to the given expression.
A) \(x + 5 = 8\)

\(x + 5 - 5 = 8 - 5\), subtract 5 from both sides of the equation

\(x = 3\), simplify and solve for x

Substitute \(x\) by 3 (solution found above) in both sides of the given equation:

Right side: \(3 + 5 = 8\)

Left side = \(8\)

\(x = 3\) is the solution to the given equation.

B) \(2x = 4\)

\(\frac{2x}{2} = \frac{4}{2}\), divide both sides by 2

\(x = 2\), simplify and solve for x

Substitute \(x\) by 2 (solution found above) in both sides of the given equation:

Right side: \(2(2) = 4\)

Left side = \(4\)

\(x = 2\) is the solution to the given equation.

C) \(\frac{x}{3} = 2\)

\(3(\frac{x}{3}) = 3(2)\), multiply both sides by 3

\(x = 6\), simplify and solve for x

Substitute \(x\) by 6 (solution found above) in both sides of the given equation:

Right side: \(\frac{6}{3} = 2\)

Left side = \(2\)

\(x = 6\) is the solution to the given equation.

Solutions

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