Finding the Unknown Class 7 Notes Maths Part 2 Chapter 7

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Class 7 Maths Chapter 7 Finding the Unknown Notes

Class 7 Finding the Unknown Notes

Find The Unknowns Class 7 Notes

An equation states that two things are equal, using mathematical symbols.
An equal sign (=) is used.
A variable is something that can vary. It assumes different numerical values; its value is not fixed.
These are usually denoted by letters of the English alphabet, such as x, y, z, l, m, n, p, etc.
From variables, we form an expression by operating on addition, subtraction, multiplication, and division on them.

Solving Equations Systematically Class 7 Notes

For any balanced numerical equation, if we either:

• Add the same number to both sides,
• or subtract the same number from both sides,
• or multiply by the same number to both sides,
• Or divide by the same number on both its sides, and the balance is undisturbed.

More Equations
Transposing means moving to the other side. It has the same effect as adding the same number to (or subtracting the same number from) both sides of the equation.
When we transpose a number from one side of the equation to the other side, we change its sign.

Linear Equations
A. Trial and error method
Example: Solve 2x + 3 = 9
For x = 1
LHS = 2 × 1 + 3 = 5 ≠ RHS
For x = 2
LHS = 2 × 2 + 3 = 7 ≠ RHS
For x = 3
LHS = 2 × 3 + 3 = 6 + 3 = 9 = RHS
So x = 3 is a solution of the equation.

B. Balancing Equation
Example: Solve 2x + 3 = 9
Subtract 3 from both sides
2x + 3 – 3 = 9 – 3
2x = 6
Divide both sides by 2
2x ÷ 2 = 6 ÷ 2
x = 3
So, x = 3 is a solution of the equation.

Rules for solving by transposing
Variable terms or unknown terms should be on one side and constants or numbers on the other side.
For this, we may need to add or subtract the terms and constants.
Example: 5x – 8 = 2x + 1
Subtract 2x from both sides (Bringing unknown terms to one side)
5x – 8 – 2x = 2x + 1 – 2x
3x – 8 = +1
Add 8 on both sides
3x – 8 + 8 = +1 + 8
3x = 9
To separate the coefficient from the unknown term, divide by the coefficient
3x ÷ 3 = 9 ÷ 3
x = 3
∴ x = 3 is a solution of 5x – 8 = 2x + 1.

Matchstick Pattern

First pattern 3 matchsticks
Second pattern 5 matchsticks
Third pattern 7 matchsticks
We note that 2 more matchsticks make the next pattern.
So the pattern is observed as
First pattern 3 = 1 + 2
Second pattern 5 = 1 + 2 + 2
Third pattern 7 = 1 + 2 + 2 + 2
.
.
.
For nth position matchsticks needed = 1 + 2 + 2 + … (n times) = 1 + 2n
Let 99 matchsticks be used for an n-pattern
2n + 1 = 99
Subtract 1 from both sides
⇒ 2n + 1 – 1 = 99 – 1
⇒ 2n = 98
⇒ n = 98 ÷ 2
⇒ n = 49The
49th pattern has 99 matchsticks
Let if possible y arrangements use 200 sticks.
2y + 1 = 200
Subtract 1 from both sides
⇒ 2y + 1 – 1 = 200 – 1
⇒ 2y = 199
⇒ y = 199 ÷ 2
⇒ y = 99.5
As the number of arrangements cannot be in decimal, an exact 200 matchsticks cannot be used.

C. Transposition
A term added on one side when transposed, i.e., shifted to the other side, will be subtracted. A term subtracted from one side when transposed will be added. A term multiplied on one side of the transposition will be divided, and vice versa.

Example: Solve 6y + 7 = 4y + 21
Here, 6y + 7 = 4y + 21
+4y when transposed to the Left Hand Side (LHS) will become -4y.
6y + 7 – 4y = 21
2y + 7 = 21
+7 when transposed to Right right-hand side (RHS) will become -7.
2y = 21 – 7
2y = 14
2 multiplied on Left left-hand side (LHS) will divide when transposed
y = 14 ÷ 2
y = 7

Example: \(\frac {4u}{7}\) = 16
Divisor 7, when transposed, is multiplied
4u = 16 × 7
Multiplicand 4, when transposed, becomes the divisor
u = \(\frac{16 \times 7}{4}\)
u = 28

Mind The Mistake, Mend The Mistake Class 7 Notes

The following are some equations, along with the steps used to solve them to find the value of the letter-number. Review each solution and determine whether the steps are correct. If there is a mistake, describe the mistake, correct i,t and solve the equation.

1. 4x + 6 = 10
⇒ 4x = 10 + 6 → Error
⇒ 4x = 16
⇒ x = 4
Correction
4x + 6 = 10
Subtract 6 from both sides
4x + 6 – 6 = 10 – 6
⇒ 4x = 10 – 6
⇒ 4x = 4
⇒ x = 1

2. 7 – 8z = 5
⇒ 8z = 7 – 5
⇒ 8z = 2
⇒ z = 4 → Error
Correction
7 – 8z = 5
⇒ 8z = 7 – 5
⇒ 8z = 2
Divide both sides by 8
\(\frac{8 z}{8}=\frac{2}{8}\)
⇒ z = \(\frac {1}{4}\)

3. 2v – 4 = 6
⇒ v – 4 = 6 – 2 → Error
⇒ v – 4 = 4
⇒ v = 8
Correction
2v – 4 = 6
Add 4 to both sides
2v – 4 + 4 = 6 + 4
2v = 10
Divide both sides by 2
\(\frac{2 v}{2}=\frac{10}{2}\)
⇒ v = 5

4. 5z + 2 = 3z – 4
⇒ 5z + 3z = -4 + 2 → Error
⇒ 8z = -2
⇒ z = \(-\frac {2}{8}\)
Correction
5z + 2 = 3z – 4
⇒ 5z – 3z = -4 – 2
⇒ 2z = -6
Divide both sides by 2
⇒ z = -3

5. 15w – 4w = 26
⇒ 15w = 26 + 4w → Error
⇒ 15w = 30
⇒ w = 2
Correction
15w – 4w = 26
⇒ 11w = 26
⇒ w = \(\frac {26}{11}\) = 2\(\frac {4}{11}\)

6. 3(x + 1) = -12
⇒ x + 1 = -4 → Error
⇒ x = -5
Correction
3x + 1 = -12
⇒ 3x = -12 – 1
⇒ 3x = -13
⇒ x = \(-\frac {13}{3}\) = -4\(\frac {1}{3}\)

7. 4(4q + 2) = 50
⇒ 4(4q) = 50 – 2 → Error
⇒ 16q = 48
⇒ q = 3
Correction
4(4q + 2) = 50 [Distributive Property]
⇒ 4(4q) + 4(2) = 50
⇒ 16q + 8 = 50
⇒ 16q = 50 – 8
⇒ 16q = 42
⇒ q = \(\frac {42}{16}\)
⇒ q = \(\frac {21}{8}\)
⇒ q = 2\(\frac {5}{8}\)

8. -2(3 – 4x) = 14
⇒ -6 – 8x = 14
⇒ -8x = 14 + 6
⇒ -8x = 20
⇒ x = \(\frac {-20}{8}\)
Correction
-2(3 – 4x) = 14
⇒ -6 + 8x = 14
⇒ 8x = 14 + 6
⇒ 8x = 20
⇒ x = \(\frac {20}{8}\)
⇒ x = \(\frac {5}{2}\)
⇒ x = 2\(\frac {1}{2}\)

9. 3(7y + 4) = 9 + 5y
⇒ 7y + 4 = \(\frac {9}{3}\) + 5y → Error
⇒ 7y + 4 = 3 + 5y
⇒ 7y – 5y + 4 = 3
⇒ 2y = 4 – 3 → Error
⇒ y = \(\frac {1}{2}\)
Correction
3(7y + 4) = 9 + 5y
⇒ 21y + 12 = 9 + 5y [Distributive Property]
⇒ 21y – 57 = 9 – 12
⇒ 16y = -3
⇒ y = \(-\frac {3}{16}\)

An algebraic equation is a mathematical statement that indicates the equality of two algebraic expressions. When the same operation is performed on both sides of an equation, equality is maintained. Finding a solution to an equation means finding the values of the unknowns in the expressions such that the LHS is equal to the RHS. Equations can often be solved by performing the same operation on both sides so that the value of the unknown becomes evident.

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