Basic rules for manipulating equations

1 Adding the same quantity to, or subtracting the same quantity from, both sides of an equation does not change the equality.

Thus if we subtract 12 from both sides of the equation x + 12 = 20, then x + 12 - 12 = 20 - 12 and so x = 20 - 12 = 8. We might consider this rule as being: when you move a quantity from one side of the equals sign to the other side you change its sign. This is termed transposition.

2 Multiplying, or dividing, both sides of an equation by the same non-zero quantity does not change the equality.

If we have the equation:

2x 3 = 1

4

then we can multiply both sides of the equation by 3 to give:

3 x 2x

3 = 3 x 1 4

and so 2x = ₄ ³

We can then divide both sides of the equation by 2 to give:

x = 1 2 3 4 = ³ 8

You might like to think of this rule as being: when you move a numerator from one side of an equation to the other, it becomes a denominator; when you move a denominator from one side of an equation to the other, it becomes a numerator. This is termed transposition.

In general, whatever mathematical operation we do to one side of an equation, provided we do the same to the other side of the equation then the balance is not affected. Thus we can manipulate equations without affecting their balance by, for example:

1 Adding the same quantity to both sides of the equation.

2 Subtracting the same quantity from both sides of the equation.

3 Multiplying both sides of the equation by the same quantity.

4 Dividing both sides of the equation by the same quantity.

5 Squaring both sides of the equation.

6 Taking the square roots of both sides of the equation.

7 Taking logarithms of both sides of the equation.

The term transposition is used when a quantity is moved from one side of an equation to the other side. The following are basic rules for use with transposition:

1 A quantity which is added on the left hand side of an equation side becomes subtracted on the right hand side.

2 A quantity which is subtracted on the left hand side of an equation becomes added on the right hand side.

3 A quantity which is multiplying on the right-hand side of an equation becomes a dividing quantity on the left-hand side.

4 A quantity which is dividing on the left-hand side of an equation becomes a multiplying quantity on the right-hand side.

The following examples illustrate some of these manipulations.

Example

Solve the equation 2x - 4 = x + 1.

If we add 4 to each side of the equation then:

2 x - 4 + 4 = x + 1 + 4

and so 2x = x + 5. If we now subtract x from both sides of the equation:

2x - x = x + 5 - x

and so we have x = 5 as the solution.

Alternatively we might think of the equation manipulation as: we move - 4 from the left side to the right to give:

2x = x + l + 4 = x + 5

We then move the x from the left side to the right to give:

2x - x = 5

and so we have x = 5 as the solution.

We can check this result by replacing x in the original equation by this value and confirming that the equation balances: 2 x 5 - 4 = 5 + 1.

Example

Solve the equation ² 3 x = 8.

Multiplying both sides of the equation by 3 gives:

3 x 2

3 x = 3 x 8

and so 2x = 24. Dividing both sides of the equation by 2 gives:

2x 2 = ²⁴ ₂

and so the solution is x = 12.

Alternatively we might think of the equation manipulation as: move the denominator 3 from the left side to the right to give a numerator:

2x = 3 x 8 = 24

Move the numerator 2 from the left side to the right side to give a denominator:

x = ²⁴ 2 = 12

We can check the result by replacing x in the original equation by this value. Then we have x 12 = 8.

Example

The voltage V across a resistance R is related to the current through it by the equation V = IR. When V = 2 V then I = 0.1 A. What is the value of the resistance?

Writing the equation with the numbers substituted gives 2 = 0.1R.

Multiplying both sides of the equation by 10 gives 2 x 10 = 10 x 0.1R.

Hence 20 = 1R and so the resistance R is 20 Ω. Alternatively we might consider the equation manipulation as: moving the numerator 0.1 from the right side to the left gives 2/0.1 = R and so R is 20 Ω. We can check the result by putting this value back in the original equation and confirm that it still balances: 2 = 0.1 x 20.

Example

Solve the equation 2 =

√

^L

10 .

Squaring both sides of the equation gives is the same as multiplying both sides of the equation by the same quantity since 2 is the same as

√(L/10):

2 x 2 =

√ L 10 x

√ L 10 4 = L

10

Hence, we have L = 40. We can check this result by putting the value in the original equation to give 2 = √(40/10).

Example

Solve the equation 100 = R2 + 75.

Subtracting 75 from each side of the equation gives 25 = R2. Taking the square root of both sides of the equation then gives R = 5. We can check this result by putting this value in the original equation to give

100 = 52 + 75.

Revision

1 Solve the following equations:

(a) x + 7 = 12, (b) 2x + 3 = x + 5, (c) x + 3 = 4x, (d) ½x = 12, (e) x = 4, (f) ½x + 2 = x + l, (g) 2x

5 = 12 15 , (h)

2x 5 =

12 15 , (i) x + 4 = ½, (j) 2x = 5

3 , (k) 20 = 5 √x, (l) 4 + x² = 20.

2 (a) The voltage V across a resistance R is related to the current through it by the equation V = IR. When V = 10 V then I = 2 A. What is the value of the resistance? The resistance unit will be Ω..

(b) The extension x of a spring is related to the stretching force F by the equation F = kx, where k is a constant. When F = 10 N then x = 20 mm.

What is the value of k? The unit of k will be N/mm.

(c) The pressure p exerted by a force F applied over an area A is given by the equation p = F/A. When p = 1000 N/m2 and A = 0.01 m2, what is the value of F? The unit of F will be N.

(d) The impedance Z of an electrical circuit is given by Z2 = R2 + X2. With Z = 10 Ω and R = 5 Ω, what is the value of X?

3.2.2 Brackets

Brackets are used to show terms are grouped together, e.g. 2(x + 3) indicates that we must regard the x + 3 as a single term which is multiplied

by 2. Thus when removing brackets, each term within the bracket is multiplied by the quantity outside the bracket, e.g.

2(x + 3 ) = 2x + 2 X 3 .

1 Adding and subtracting bracketed quantities

When a bracket has a + sign in front of the bracket then effectively we are multiplying all the terms in the bracket by +1, e.g.

a + (b + c) = a + b + c

When a bracket has a - sign in front of the bracket then we are multiplying all the terms in the bracket by - 1 , e.g.

a - (b + c) = a - b - c

2 Multiplication of two bracketed quantities

Consider the removal of brackets when we have (a + b)(c + d).

Following the above rule for the removal of brackets, each term within the first bracket must be multiplied by quantity inside the second bracket to give:

a(c + d) + b(c + d) = ac + ad + bc + bd

Note that the resulting expression is given by multiplying each term in one bracket by each term in the other bracket.

3 Square of bracket quantities

Another example of multiplication is the square of a bracketed term:

(a + b)² = (a + b)(a + b) = a(a + b) + b(a + b)

= a² + ab + ab + b² = a² + 2ab + b²

The square is the sum of the squares of the two terms and twice their product.

4 Product of the sum and difference of two terms

Another example of multiplication is the product of the sum and difference of two terms:

(a + b)(a - b) = a(a - b) + b(a - b) = a2 - ab + ab - b2

= a2 - b2

The result is the difference of the squares of the two terms.

Example

Remove the brackets in the following:

(a) 5(x + 2), (b) 2 - 3(x + 5), (c) 2(x + 2) + 3(x - 4), (d) (x + l)(x + 3), (e) (x + 5)², (f) (x + 5)(x - 5), (g) (2x + 1)(5x - 2).

(a) 5(x + 2) = 5x + 10

(b) 2 - 3(x + 5) = 2 - 3x - 15 = -3x - 13

(c) 2(x + 2) + 3(x - 4) = 2x + 4 + 3x - 12 = 5x - 8

(d) (x + 1)(x + 3) = x(x + 3) + 1(x + 3) = x2 + 3x + x + 3 = x2 + 4x + 3 (e) (x + 5)2 = (x + 5)(x + 5) = x(x + 5) + 5(x + 5) = x2 + 5x + 5x + 25

= x² + 10x + 25

(f) (x + 5)(x - 5) = x(x - 5) + 5(x - 5) = x² - 5x + 5x - 25 = x² - 25 (g) (2x + 1)(5x - 2) = 2x(5x - 2) + 1 ( 5 x - 2 ) = 10x2 - 4x + 5x - 2

= 10x2 + x - 2 Example

Solve the equation 2x + 3

3 = x + 1 4

Multiplying both sides of the equation by 3 gives:

3

(

^{2x + 3} ₃

)

^{= 3}

(

^{x + 1} ₄

)

2x + 3 = 3

(

^{x + 1} 4

)

Moving the numerator 4 from the right side to left gives:

4(2x + 3) = 3(x + 1)

Removing the brackets gives 8x + 12 = 3x + 3 and hence 5x = - 9 and so x = -9/5.

Multiplying both sides of the equation by 4 gives 4(2x + 3) = 4 x 3

(

^{x + 1} 4

)

= 3(x + l)

and so 8x + 12 = 3x + 3. Subtracting 12 from both sides of the equation gives 8x + 12 - 12 = 3x + 3 - 12 and so 8x = 3x - 9. Subtracting 3x from both sides of the equation gives

8x - 3x = 3x - 3x - 9 Hence 5x = - 9 and so x = -9/5.

Alternatively we could have tackled the equation by transposing numbers from side of the equals sign to the other. Thus, moving the numerator 3 from the left side to right gives:

2x + 3 = 3

(

^{x + 1} _{4 )}

Revision

3 Remove the brackets in the following:

(a) 5 + (2x + 1), (b) 4 - (x + 3), (c) 2(x + 1) + 4(x + 5), (d) 5(2x - 1) + 3(x + 7), (e) (x + 3)(2x + 7), (f) (x - 3)²,

(g) (2x + 1)², (h) (x + 1)(x - 1), (i) x(2x + 1), (j) x(x + 2) + 3(x + 5).

In document Mathematics for Engineering (Page 42-48)