Find the rational number such that:

Understand the Question We are asked to find the rational number that satisfies the given linear equation. By applying the distributive property to the left-hand side (LHS), we can simplify the expression. If both sides reduce to the same expression (an identity), the equation holds true for any rational number . Step 1 · Simplify the Left Hand Side Using the distributive property Step 2 · Compare LHS and RHS to Find Substitute the simplified LHS back into the equation Subtract from both sides Since the equation simplifies to the true statement independently of , it is an identity. Answer > can be any rational number Common Mistakes Assuming or No Solution: When the variable terms cancel out and leave a true statement (like ), it means the equation has infinitely many solutions (any rational number), not or "no solution". Distribution Error: Forgetting to multiply with the second term inside the parentheses .

The World of Numbers | Exercise 3.3

Question 8

Find the rational number $x$ such that:

$\dfrac{5}{6}\left(x + \dfrac{3}{5}\right) = \dfrac{5}{6}x + \dfrac{1}{2}$

Solution

Understand the Question

We are asked to find the rational number $x$ that satisfies the given linear equation.
By applying the distributive property $a(b + c) = ab + ac$ to the left-hand side (LHS), we can simplify the expression.
If both sides reduce to the same expression (an identity), the equation holds true for any rational number $x$ .

Step 1 · Simplify the Left Hand Side

Using the distributive property $a(b + c) = ab + ac$

\begin{aligned} \dfrac{5}{6}\left(x + \dfrac{3}{5}\right) &= \dfrac{5}{6}x + \dfrac{5}{6} \times \dfrac{3}{5} \\[0.6em] &= \dfrac{5}{6}x + \dfrac{5 \times 3}{6 \times 5} \\[0.6em] &= \dfrac{5}{6}x + \dfrac{15}{30} \\[0.6em] &= \dfrac{5}{6}x + \dfrac{1}{2} \end{aligned}

Step 2 · Compare LHS and RHS to Find $x$

Substitute the simplified LHS back into the equation $\dfrac{5}{6}x + \dfrac{1}{2} = \dfrac{5}{6}x + \dfrac{1}{2}$

Subtract $\dfrac{5}{6}x$ from both sides

\begin{aligned} \dfrac{5}{6}x + \dfrac{1}{2} - \dfrac{5}{6}x &= \dfrac{5}{6}x + \dfrac{1}{2} - \dfrac{5}{6}x \\[0.6em] \dfrac{1}{2} &= \dfrac{1}{2} \end{aligned}

Since the equation simplifies to the true statement $\dfrac{1}{2} = \dfrac{1}{2}$ independently of $x$ , it is an identity.

Answer

$x$ can be any rational number

Common Mistakes

Assuming $x = 0$ or No Solution: When the variable terms cancel out and leave a true statement (like $\dfrac{1}{2} = \dfrac{1}{2}$ ), it means the equation has infinitely many solutions (any rational number), not $x = 0$ or "no solution".
Distribution Error: Forgetting to multiply $\dfrac{5}{6}$ with the second term inside the parentheses $\left(\dfrac{3}{5}\right)$ .

Question 8

More questions in Exercise 3.3