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NCERT Solutions for Class 7th Maths Chapter 7 .2 Solving Equations Systematically — In-text Questions

Book page 170 – 1727 Updated on2026-09-19

Q1.
Let us use these ideas to solve the equation 5x – 4 = 7. What can we do so that 5x is on one side and the equality between the LHS and the RHS still holds?
Answer

Add 4 to both sides. Then x = 11/5.

5x − 4 = 7
To keep only 5x on the LHS, remove the term − 4 by adding 4 to both sides:
5x − 4 + 4 = 7 + 4
5x = 11

To keep only x on the LHS, remove the factor 5 by dividing both sides by 5:
5x ÷ 5 = 11 ÷ 5
x = 11/5
Why it happens: two different kinds of clutter sit around the x. The − 4 is a term, so it is removed by adding its additive inverse, + 4. The 5 is a factor, so it is removed by dividing. Deal with the term first: if you divide by 5 at the start you get x − 4/5 = 7/5, which works but brings in fractions sooner.
Tip: "add 4 to both sides" and "take the − 4 across and change its sign" are the same move. The first version explains why the second is allowed.
Q2.
Can we check that x = 11/5 is the correct solution to the equation 5x – 4 = 7?
Answer

Yes — substitute 11/5 for x and see that LHS = RHS.

LHS = 5 × (11/5) − 4
= 11 − 4  (the 5 and the 5 cancel)
= 7

RHS = 7
LHS = RHS, so x = 11/5 is indeed the solution.
Why it happens: solving an equation is a chain of steps, and any one of them could hide a slip. Substituting the answer back tests the original equation, not your working, so it catches mistakes that re-reading would miss.
Tip: make checking a habit. It takes one line and it is the only way to be sure.
Q3.
Example 5: Solve the equation 11y + (–5) = 61. Can you check that this solution is correct?
Answer

y = 6, and the check confirms it.

11y + (−5) = 61
Remove the term (−5) by subtracting (−5) from both sides:
11y + (−5) − (−5) = 61 − (−5)
11y = 61 + 5 = 66

Now remove the factor 11 by dividing both sides by 11:
11y ÷ 11 = 66 ÷ 11
y = 6
Check: LHS = 11 × 6 + (−5) = 66 − 5 = 61 = RHS ✓
Why it happens: subtracting (−5) is the same as adding 5, because the additive inverse of −5 is +5. That is why the 61 grows to 66 rather than shrinking. Once 11y = 66, you can either divide by 11 or simply recall that 11 × 6 = 66 — both are the same step.
Tip: when a negative number appears in brackets, write the subtraction out fully before simplifying. 61 − (−5) = 61 + 5 is where most slips happen.
Q4.
Example 6: Solve 6y + 7 = 4y + 21. In this equation, expressions with an unknown are on both sides. What can be done to bring the unknown terms to the same side?
Answer

Subtract 4y from both sides. Then y = 7.

6y + 7 = 4y + 21
Subtract 4y from both sides:
6y + 7 − 4y = 4y + 21 − 4y
2y + 7 = 21

Subtract 7 from both sides:
2y + 7 − 7 = 21 − 7
2y = 14

Divide both sides by 2:
2y ÷ 2 = 14 ÷ 2
y = 7
Check: LHS = 6 × 7 + 7 = 49. RHS = 4 × 7 + 21 = 49. ✓
Why it happens: this is Fig. 7.10 all over again — sacks on both pans. Taking 4y off each side removes the same amount from both, so the equality holds, and it leaves the unknown on one side only. After that the equation has the familiar shape 2y + 7 = 21, which we already know how to finish.
Tip: subtract the smaller multiple of y (here 4y, not 6y). That keeps the coefficient of y positive and avoids negative numbers.
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