Grade 7 Algebra Questions and Problems with Solutions

Substitute each variable by its given numerical value, apply the order of operations, and simplify.

1. \(12 (-1)^3 + 5 (-1)^2 + 4 (-1) - 6 = 12(-1) + 5(1) - 4 - 6 = -12 + 5 - 4 - 6 = \mathbf{-17}\)
2. \(2 (2)^2 + 3 (-2)^3 - 10 = 2(4) + 3(-8) - 10 = 8 - 24 - 10 = \mathbf{-26}\)
3. \(\dfrac{-2(2) - 1}{2 + 3} = \dfrac{-4 - 1}{5} = \dfrac{-5}{5} = \mathbf{-1}\)
4. \(2 + 2|(-4) - 4| = 2 + 2|-8| = 2 + 2 \times 8 = 2 + 16 = \mathbf{18}\)

Use the distribution rule \(a(b + c) = ab + ac\) to expand the expressions, and then group like terms.

1. \(-2(x - 8) + 3(x - 7) = -2x + 16 + 3x - 21\)
   \(= (-2x + 3x) + (16 - 21) = \mathbf{x - 5}\)
2. \(2(a + 1) + 5b + 3(a + b) + 3 = 2a + 2 + 5b + 3a + 3b + 3\)
   \(= (2a + 3a) + (5b + 3b) + (2 + 3) = \mathbf{5a + 8b + 5}\)
3. \(a(b + 3) + b(a - 2) + 2a - 5b + 8 = ab + 3a + ba - 2b + 2a - 5b + 8\)
   \(= (ab + ba) + (3a + 2a) + (-2b - 5b) + 8 = \mathbf{2ab + 5a - 7b + 8}\)
4. \(\dfrac{1}{2}(4x + 4) + \dfrac{1}{3}(6x + 12) = 2x + 2 + 2x + 4\)
   \(= (2x + 2x) + (2 + 4) = \mathbf{4x + 6}\)
5. \(4(-x + 2 - 3(x - 2)) = -4x + 8 - 12x + 24\)
   \(= (-4x - 12x) + (8 + 24) = \mathbf{-16x + 32}\)

Simplify using the rules for fractions (addition, multiplication, and division).

1. \(\dfrac{x}{y} + \dfrac{4}{y} = \mathbf{\dfrac{x + 4}{y}}\)
2. \(\left(\dfrac{2x}{4}\right) \times \dfrac{1}{2} = \dfrac{2x \times 1}{4 \times 2} = \dfrac{2x}{8} = \mathbf{\dfrac{x}{4}}\)
3. \(\dfrac{3x}{5} \div \dfrac{x}{5} = \dfrac{3x}{5} \times \dfrac{5}{x} = \dfrac{15x}{5x} = \mathbf{3}\)

Use the rules of multiplication and division with exponents (add exponents when multiplying like bases, subtract when dividing).

1. \(3 x^{2} \times 5 x^{3} = (3 \times 5) x^{2 + 3} = \mathbf{15 x^{5}}\)
2. \(\dfrac{(2 y)^4 \cdot 9 x^3}{4 y^4 (3 x)^2} = \dfrac{16 y^4 \cdot 9 x^3}{4 y^4 \cdot 9 x^2} = \dfrac{16 \times 9}{4 \times 9} y^{4 - 4} x^{3 - 2} = \mathbf{4x}\)

Find the greatest common factors (GCF) and factor the expressions using distribution backwards.

1. \(9x - 3 = 3 \times 3x - 3 \times 1 = \mathbf{3(3x - 1)}\)
2. \(24x + 18y = 6 \times 4x + 6 \times 3y = \mathbf{6(4x + 3y)}\)
3. \(bx + dx = \mathbf{x(b + d)}\)

Solve equations step-by-step by performing inverse operations to isolate \(x\).

1. \[ \begin{aligned} 2x + 5 &= 11 \\ 2x &= 6 \\ x &= \mathbf{3} \end{aligned} \]
2. \[ \begin{aligned} 3x &= \dfrac{6}{5} \\ 15x &= 6 \quad \text{(multiplying both sides by 5)} \\ x &= \dfrac{6}{15} = \mathbf{\dfrac{2}{5}} \end{aligned} \]
3. \[ \begin{aligned} 3(2x + 2) + 2 &= 20 \\ 6x + 6 + 2 &= 20 \\ 6x + 8 &= 20 \\ 6x &= 12 \\ x &= \mathbf{2} \end{aligned} \]

Rewrite repeated multiplication products using exponents.

\[3 \times a \times a \times a - 5 \times b \times b = \mathbf{3 a^3 - 5 b^2}\]

Use the perimeter formula \(P = 2 \times \text{length} + 2 \times \text{width}\), substitute the expressions and the known perimeter, and solve for \(x\).

\[ \begin{aligned} 2(2x + 3) + 2(x + 1) &= 32 \\ 4x + 6 + 2x + 2 &= 32 \\ 6x + 8 &= 32 \\ 6x &= 24 \\ x &= \mathbf{4} \end{aligned} \]

Use the area formula \(A = \text{width} \times \text{length}\), substitute the known values, and solve.

\[ \begin{aligned} 3(2x - 1) &= 27 \\ 6x - 3 &= 27 \\ 6x &= 30 \\ x &= \mathbf{5} \end{aligned} \]

Since 45% are male, \( 100\% - 45\% = 55\% \) are female.

Calculate the ratio of females to males.

\[ \text{Ratio} = \dfrac{55\%}{45\%} = \dfrac{55}{45} = \mathbf{\dfrac{11}{9}} \]

Use the formula: \(\text{Distance} = \text{time} \times \text{speed}\).

\[ \begin{aligned} 300 &= 3(x + 30) \\ 300 &= 3x + 90 \\ 210 &= 3x \\ x &= \mathbf{70} \end{aligned} \]

Solve the proportion by cross-multiplication.

\[ \begin{aligned} \dfrac{4}{5} &= \dfrac{a}{16} \\ 4 \times 16 &= 5 \times a \\ 64 &= 5a \\ a &= \dfrac{64}{5} = \mathbf{12.8} \end{aligned} \]

Substitute the ordered pair \(x = 2\) and \(y = a + 2\) into the given equation and solve for \(a\).

\[ \begin{aligned} 2(2) + 2(a + 2) &= 10 \\ 4 + 2a + 4 &= 10 \\ 2a + 8 &= 10 \\ 2a &= 2 \\ a &= \mathbf{1} \end{aligned} \]

List all factors of 25 and 45.

Factors of 25 are: 1, 5, 25

Factors of 45 are: 1, 3, 5, 9, 15, 45

The greatest common factor of 25 and 45 is \(5\).

\[ \mathbf{1,234,750,002} \]

three hundred ninety-three billion, two hundred thirty-four million, thirty-four

Find the first few multiples of 15 and 35 until you get one that is common:

Multiples of 15: \(15, 30, 45, 60, 75, 90, 105, 120, 135, \ldots\)

Multiples of 35: \(35, 70, 105, 140, \ldots\)

Select the lowest common multiple (LCM):

The LCM is \(105\).

\(\dfrac{2}{3}\) of \(x\) is 30 is mathematically written as:

\[ \dfrac{2}{3} \times x = 30 \]

Multiply by 3 and simplify:

\[ 2x = 90 \]

Solve for \(x\):

\[ x = 90 \div 2 = \mathbf{45} \]

20% of \(\dfrac{1}{3}\) is written as:

\[ 20\% \times \dfrac{1}{3} \]

Use the fraction equivalent \( 20\% = \dfrac{20}{100} \) to simplify:

\[ = \dfrac{20}{100} \times \dfrac{1}{3} = \dfrac{20}{300} = \mathbf{\dfrac{1}{15}} \]

Let the smaller number be \( x \). Then the larger number is \( x + 17 \).

Their sum is given as 69, hence: \[ x + (x + 17) = 69 \]

Simplify: \[ 2x + 17 = 69 \]

Subtract 17 from both sides: \[ 2x = 52 \]

Divide both sides by 2 to solve for \( x \): \[ x = 26 \]

So the larger number is: \[ x + 17 = 26 + 17 = \mathbf{43} \]

Convert all values to decimals and order:

\[ \begin{aligned} \dfrac{12}{5} &= 2.4 \\ 250\% = \dfrac{250}{100} &= 2.5 \\ \dfrac{21}{10} &= 2.1 \\ 2.3 &= 2.3 \end{aligned} \]

Order from smallest to largest (2.1, 2.3, 2.4, 2.5):

\( \mathbf{\dfrac{21}{10}, 2.3, \dfrac{12}{5}, 250\%} \)

Let \( x \), \( x + 1 \), and \( x + 2 \) be the 3 consecutive integers.

The sum of the three consecutive integers is 96, hence:

\[ x + (x + 1) + (x + 2) = 96 \]

Group like terms:

\[ \begin{aligned} 3x + 3 &= 96 \\ 3x &= 93 \\ x &= 31 \end{aligned} \]

The largest of these numbers is \( x + 2 \):

\[ x + 2 = 31 + 2 = \mathbf{33} \]

Let \(x\) be the geography score. Chemistry is \(2x\). The average of the 4 courses is 79:

\[ \dfrac{93 + 88 + x + 2x}{4} = 79 \]

Group like terms in the numerator:

\[ \dfrac{3x + 181}{4} = 79 \]

Multiply both sides by 4 and simplify:

\[ \begin{aligned} 3x + 181 &= 316 \\ 3x &= 135 \\ x &= 45 \end{aligned} \]

His score in Geography is \(x = \mathbf{45}\).

His score in Chemistry is \(2x = 2(45) = \mathbf{90}\).

Let \(x\) be the score in English.

The score in Math is: \( x + 7 \)

The score in Physics is 5 more than Math: \( (x + 7) + 5 = x + 12 \).

Linda scored a total of 265 points:

\[ x + (x + 7) + (x + 12) = 265 \]

Group like terms and simplify:

\[ \begin{aligned} 3x + 19 &= 265 \\ 3x &= 246 \\ x &= 246 \div 3 = 82 \end{aligned} \]

Score in English is \( x = \mathbf{82} \)

Score in Math is \( x + 7 = 82 + 7 = \mathbf{89} \)

Score in Physics is \( x + 12 = 82 + 12 = \mathbf{94} \)

Each bicycle has 2 wheels, hence the number of bicycles is:

\[ \dfrac{100}{2} = \mathbf{50} \text{ bicycles} \]

The number of Car wheels is the total minus the bicycle wheels:

\[ 300 - 100 = 200 \text{ car wheels} \]

Each car has 4 wheels, hence the number of cars is:

\[ \dfrac{200}{4} = \mathbf{50} \text{ cars} \]