iGCSE Mathematics (0580) : C2.5 Derive and solve simple linear equations in one unknown.iGCSE Style Questions Paper 1

Question

Solve the following equations.
(a) x + 9 = 16

(b) 6y = 27

▶️Answer/Explanation

(a) To solve the equation \(x + 9 = 16\), we can isolate \(x\) by subtracting 9 from both sides of the equation:
\(x + 9 – 9 = 16 – 9\)
\(x = 7\)
Therefore, the solution to the equation is \(x = 7\).
(b) To solve the equation \(6y = 27\), we can isolate \(y\) by dividing both sides of the equation by 6:
\(\frac{6y}{6} = \frac{27}{6}\)
\(y = \frac{27}{6}\)
\(y = \frac{9}{2}\)
Therefore, the solution to the equation is \(y = \frac{9}{2}\) or \(y = 4.5\).

 

Question

Solve the equation 4x – 2 = 7 .

▶️Answer/Explanation

To solve the equation \(4x – 2 = 7\), we can isolate \(x\) by performing the following steps:
\(4x – 2 + 2 = 7 + 2\)
\(4x = 9\)
\(\frac{4x}{4} = \frac{9}{4}\)
\(x = \frac{9}{4}\)
Therefore, the solution to the equation is \(x = \frac{9}{4}\) or \(x = 2.25\).

Question

w = 3a – 5b
Calculate w when a = 2 and b = –3.

▶️Answer/Explanation

To calculate the value of \(w\) when \(a = 2\) and \(b = -3\) in the equation \(w = 3a – 5b\), we can substitute the given values and evaluate the expression:
\(w= 3(2) – 5(-3)\)
Simplifying:
\(w = 6 + 15\)
\(w = 21\)
Therefore, when \(a = 2\) and \(b = -3\), the value of \(w\) is 21.

Question

 Solve the equation.
5 – 2x = 3x – 19

▶️Answer/Explanation

To solve the equation \(5 – 2x = 3x – 19\),
\(5 – 2x + 2x = 3x – 19 + 2x\)
\(5 = 5x – 19\)
\(5 + 19 = 5x – 19 + 19\)
\(24 = 5x\)
\(\frac{24}{5} = \frac{5x}{5}\)
\(x = \frac{24}{5}\)
Therefore, the solution to the equation is \(x = \frac{24}{5}\), which can be expressed as a decimal as \(x = 4.8\).

 

Question

 (a) Multiply out the brackets.
5(x + 3)
(b) Factorise completely.
\(12xy – 3x^2\)
(c) Solve:
5x – 24 = 51

▶️Answer/Explanation

(a) To multiply out the brackets 5(x + 3), you distribute the 5 to each term inside the brackets:
\(5(x + 3) = 5\times x + 5\times 3 = 5x + 15\)
(b) To factorize completely the expression 12xy – 3x^2, you can factor out the greatest common factor, which is 3x:
12xy – 3x^2 = 3x(4y – x)
(c) To solve the equation 5x – 24 = 51, you can isolate the variable x by performing inverse operations. First, add 24 to both sides to get rid of the constant term on the left side:
5x – 24 + 24 = 51 + 24
5x = 75
\(\frac{5x}{5}=\frac{75}{5}\)
x = 15
Therefore, the solution to the equation 5x – 24 = 51 is x = 15.

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