Practice Problems (Junior Cert Mathematics): Revision Notes

Practice Problems

Questions:

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Problem 1

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Problem 2

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Problem 3

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Problem 4

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Problem 5

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Solutions:

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Problem 1

Step 1: Isolate the variable term.

• Subtract $7$ from both sides to get the term with $x$ by itself. $3x + 7 - 7 = 22 - 7$ Simplifying: $3x = 15$ Step 2: Solve for $x$.

• Divide both sides by $3$ to solve for $x$. $\frac{3x}{3} = \frac{15}{3}$ Simplifying: $x = 5$ Solution: The solution is $x = 5$.

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Problem 2

Step 1: Get all the $x$ terms on one side.

• Subtract $3x$ from both sides to move the $x$ terms together. $5x - 3x - 9 = 3x - 3x + 11$ Simplifying: $2x - 9 = 11$ Step 2: Isolate the variable term.

• Add 9 to both sides to get the $x$ term by itself. $2x - 9 + 9 = 11 + 9$ Simplifying: $2x = 20$ Step 3: Solve for $x$.

• Divide both sides by $2$ to solve for $x$. $\frac{2x}{2} = \frac{20}{2}$ Simplifying: $x = 10$ Solution: The solution is $x = 10$.

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Problem 3

Step 1: Clear the fractions.

• Multiply both sides by $4$, the least common multiple of the denominators, to eliminate the fractions. $4 \times \frac{2x - 3}{4} = 4 \times \frac{5x + 1}{2}$ Simplifying: $2x - 3 = 2(5x + 1)$ Step 2: Distribute and simplify.

• Distribute the $2$ on the right side: $2x - 3 = 10x + 2$ Step 3: Get all the $x$ terms on one side.

• Subtract $2x$ from both sides: $2x - 2x - 3 = 10x - 2x + 2$ Simplifying: $-3 = 8x + 2$ Step 4: Isolate the variable term.

• Subtract $2$ from both sides: $-3 - 2 = 8x + 2 - 2$ Simplifying: $-5 = 8x$ Step 5: Solve for $x$.

• Divide both sides by 8 to solve for $x$. $\frac{-5}{8} = x$ Solution: The solution is $x = \frac{-5}{8}$.

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Problem 4

Step 1: Factorise the quadratic expression.

• We need to find two numbers that multiply to $+6$ (the constant term) and add to $-5$ (the coefficient of $x$).

• The numbers $-2$ and $-3$ work because:

  • $(-2) \times (-3) = +6$
  • $(-2) + (-3) = -5$

• Factorising the quadratic expression: $(x - 2)(x - 3) = 0$ Step 2: Solve for $x$.

• Set each factor equal to $0$: $x - 2 = 0 \quad \text{or} \quad x - 3 = 0$

• Solve each equation: $x = 2 \quad \text{or} \quad x = 3$ Solution: The solutions are $x = 2$ and $x = 3$.

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Problem 5

Step 1: Identify the coefficients.

• For the quadratic equation $ax^2 + bx + c = 0$, identify the coefficients:

  • $a = 2$
  • $b = 3$
  • $c = -2$ Step 2: Write down the quadratic formula.

• The quadratic formula is: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Step 3: Substitute the values into the formula. $x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}$ Simplifying: $x = \frac{-3 \pm \sqrt{9 + 16}}{4}$ $x = \frac{-3 \pm \sqrt{25}}{4}$ $x = \frac{-3 \pm 5}{4}$

Step 4: Solve for the two possible values of $x$.

• First solution: $x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$
• Second solution: $x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$ Solution: The solutions are $x = \frac{1}{2}$ and $x = -2$.

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