Equations

Chapter 38

5 Steps - 3 Clicks

Equations

Introduction

To assess a mathematical expression/equation, understand the given statements in the question and compute/solve the expression as per the formulae/rules for the order of operations.

Methods

By comparing two mathematical expressions by utilizing the indication of equality, an equation is framed. An equation may contain one or more variables. If the equation has only one variable then the equation itself is adequate to acquire the estimation of the variable. If the equation has two variables then two consistent conditions/equations are required to get the estimation of the variables. As a rule, an equation has ‘n’ variables then ‘n’ predictable conditions/equations are required to get all the estimations of the ‘n’ variables. So, when there are ‘n’ number of unknowns, we need ‘n’ number of equations to solve for the values of ‘n’ unknowns. For example: If there are 2 variables x and y, we need 2 equations to solve for the values of x and y.

Each arithmetical expression can be composed as a single term or a progression of terms separated by plus or minus signs.

For example:

• The expression 3$$x^2$$ + 10 has two terms;

• The expression 1.14S is a single term;

• The expression $$\frac{z – 1}{4}$$, which can be written as $$\frac{z}{4} – \frac{1}{4}$$, has two terms;

• In the expression: 2$$x^2$$ + 7$$x$$ – 5, 2 is the coefficient of the $$x^2$$ term, 7 is the coefficient of the $$x$$ term, and 5 is the constant term.

A statement that equates two algebraic expressions is called an equation.

There are three types of equations. They are:

1. Linear equation in one variable.

Example: 3$$x$$ + 5 = -2

2. Linear equation in two variables.

Example: $$x$$ -3$$y$$ = 10

3. Quadratic equation in one variable.

Example: 20$$y^2$$ + 6$$y$$ -15 = 0

1.Solving equations with one variable:-

To solve a linear equation in one variable intends to discover the estimation of the variable that makes the equation true. Two equations that have the same solution are said to be equivalent.

For example,

$$x$$ + 2 = 4 and 4$$x$$ + 1 = 2 are equivalent equations

Here, both are true when $$x$$ = 1 and are false otherwise.

Two basic rules are important for solving linear equations:-

(i) When the same constant is added to (or subtracted from) both sides of an equation, the equality is preserved, and the new equation is equivalent to the original.

(ii) When both sides of an equation are multiplied (or divided) by the same non-zero constant, the equality is preserved, and the new equation is equivalent to the original.

Example 1:

5$$x$$ – 2 = 13

Solution:

Add 2 on both sides of equations, i.e.

5$$x$$ – 2 + 2 = 13 + 2

5$$x$$ = 15

$$x$$ = 3

Example 2:
Solve the equation: 4$$x$$ = 8

Solution:

In this example, the 4 is multiplying the $$x$$. Therefore, to isolate $$x$$, you must divide that side by 4. When doing this, you must remember one important rule: whatever you do to one side of the equation, you must do to the other side. So we will divide both sides by 4.

4$$x$$ = 8

$$\frac{4x}{4}$$ = $$\frac{8}{4}$$

Simplifying:

$$x$$ = 2

Example 3:

Solve: 3x=12

Solution:

Since $$x$$ is being multiplied by 3, the plan is to divide by 3 on both sides:

3$$x$$ = 12

$$\frac{3x}{3}$$ = $$\frac{12}{3}$$

$$x$$ = 4

2. Solving equations with two variables:-

To calculate linear equations in two variables, it is important to have two equations that are not proportionate. To explain such a “system” of concurrent equations, there are two basic methods.

• In the first method, either equations can be utilized to express one variable as far as the other.

Example: Solve the two equations, 4$$x$$ + 3$$y$$ = 13 and $$x$$ + 2$$y$$ = 2

4$$x$$ + 3$$y$$ = 13 ——-(i)

$$x$$ + 2$$y$$ = 2 ——-(ii)

Substitute (ii) in (i)

4(2 – 2$$y$$) + 3$$y$$ = 13

8 – 8$$y$$ + 3$$y$$ = 13

$$y$$ = -1

Now, substitute the value of $$y$$ in (ii)

$$x$$ + 2(-1) = 2

$$x$$ = 4

• In the second method, the object is to make the coefficients of one variable the same in both equations so that one variable can be disposed of by either including both equations together or subtracting one from the other.

Example: Solve the two equations, 4$$x$$ + 3$$y$$ = 13 and 4$$x$$ + 8$$y$$ = 8

4$$x$$ + 3$$y$$ = 13 ——(i)

4$$x$$ + 8$$y$$ = 8 ——(ii)

Subtract (ii) from (i), then

-5$$y$$ = 5

$$y$$ = -1

Substitute $$y$$ value in (i), then

$$x$$ = -1

3. Solving quadratic equations with one variable:-

A quadratic equation is any equation that can be expressed in the form of a$$x^2$$ + b$$x$$ + c = 0.

Where a, b,and c are real numbers (a ≠ 0).

Such an equation can be solved by the formula

$$x$$ = $$\frac{-b ± \sqrt{b^2 – 4ac}}{2a}$$

Example:

$$x^2$$ + 8$$x$$ + 15 = 0

($$x$$ + 3)($$x$$ + 5) = 0

Therefore, $$x$$ + 3 = 0

$$x$$ = -3

$$x$$ + 5 = 0

$$x$$ = -5

Formulae

1. $$x^{-a}$$ = $$\frac{1}{x^{a}}$$ ($$x$$ ≠ 0)

2. $$(x^a)(x^b)$$ = $$x^{a + b}$$

3. $$(x^a)(y^a)$$ = $$(xy)^a$$

4. $$\frac{x^a}{x^b}$$ = $$x^{a – b}$$ = $$\frac{1}{x^{a – b}}$$ ($$x$$ ≠ 0)

5. $$\frac{x^a}{y^a}$$ = $$(\frac{x}{y})^a$$ ($$y$$ ≠ 0)

6. $$(x^a)^b$$ = $$x^{ab}$$

7. If $$x$$ ≠ 0, then $$x^0$$ = 1

Samples

1. Solve $$y^2$$ + 2$$y$$ – 10 = 0

Solution:

Given equation is $$y^2$$ + 2$$y$$ – 10 = 0

Here,

a = 1, b = 2 and c = -10

Consider, $$y$$ = $$\frac{-b ± \sqrt{b^2 – 4ac}}{2a}$$

$$y$$ = $$\frac{-2 ± \sqrt{2^2 – 4(1)(-10)}}{2(1)}$$

$$y$$ = $$\frac{-2 ± \sqrt{44}}{2}$$

$$y$$ = $$\frac{-2 ± \sqrt{44 * 11}}{2}$$

$$y$$ = $$\frac{-2 ± 2\sqrt{11}}{2}$$

$$y$$ = $$\frac{2(-1 ± 2\sqrt{11})}{2}$$

$$y$$ = -1 ± $$\sqrt{11}$$

Therefore, $$y$$ = -1 ± $$\sqrt{11}$$

2. Solve 3$$z^2$$ -5$$z$$ + 2 = 0

Solution:

Given equation is

3$$z^2$$ -5$$z$$ + 2 = 0

$$z^2$$ – $$\frac{5}{3}z$$ + $$\frac{2}{3}$$ = 0

Add $$\frac{1}{36}$$ on both sides

$$z^2$$ – $$\frac{5}{3}z$$ + $$\frac{2}{3}$$ + $$\frac{1}{36}$$ =
$$\frac{1}{36}$$

$$z^2$$ – $$\frac{5}{3}z$$ + $$\frac{25}{36}$$ = $$\frac{1}{36}$$

$$(z – \frac{5}{6})^2$$ = ±$$\frac{1}{6}$$

$$z$$ = $$\frac{2}{3}$$ or 1

3. A gathering of students are at visit. The aggregate toll is Rs. 120 and this is to be shared similarly among the students. In the event that two more students join the visit, each will pay Rs. 2 less. find the original number of student in the gathering.

Solution:

Let the original number of students be $$y$$

$$\frac{120}{y}$$ = $$\frac{120}{2 + y}$$ + 2

$$\frac{120}{y}$$ = $$\frac{124 + 2y}{2 + y}$$

240 + 120$$y$$ = 124$$y$$ + 2$$y^2$$

2$$y^2$$ + 4$$y$$ = 240

$$y^2$$ + 2$$y$$ = 120

$$(y+1)^2$$ = 121

$$y$$ + 1 = 11

$$y$$ = 10 or -12

Therefore, original number is 10.

4. Quantitative comparison of

$$x$$ + $$y$$ = 5

$$y$$ – $$x$$ = 3

 Quantity A Quantity B $$x$$ $$y$$

A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

Given that $$x$$ + $$y$$ = 5

$$y$$ – $$x$$ = 3

This is a pair of simultaneous equations.

By adding them together, 2$$y$$ = 8 is obtained.

Now, $$y$$ = 4.

Substituting $$y$$ value in the first equation gives $$x$$ + 4 = 5

Now, $$x$$ = 1.

The right hand column is greater.

Therefore, the correct option is B.

5. Quantitative comparison of

$$x^2$$ – 12$$x$$ + 35 = 0

 Quantity A Quantity B $$x$$ 12

A. The quantity on the left is greater.
B. The quantity on the right is greater.
C. Both are equal.
D. The relationship cannot be determined without further information.

Solution:

Given $$x^2$$ – 12$$x$$ + 35 = 0

This is a quadratic equation.

By factorize the given equation i.e.

($$x$$ – 5)($$x$$ – 7) = 0

So $$x$$ must be either 5 or 7.

Both answers are less than 12.

Therefore, option B is correct one.