To remember the process of framing simultaneous linear equations from mathematical problems

● To remember how to solve simultaneous equations by the method of comparison and method of elimination

● To acquire the ability to solve simultaneous equations by the method of substitution and method of cross-multiplication

● To know the condition for a pair of linear equations to become simultaneous equations

● To acquire the ability to solve mathematical problems framing simultaneous equations

We know that if a pair of definite values of two unknown quantities satisfies simultaneously two distinct linear equations in two variables, then those two equations are called simultaneous equations in two variables. We also know the method of framing simultaneous equations and two methods of solving these simultaneous equations.

We have already learnt that linear equation in two variable x and y is in the form ax + by + c = 0.

Where a, b, c are constant (real number) and at least one of a and b is non-zero.

The graph of linear equation ax + by + c = 0 is always a straight line.

Every linear equation in two variables has an infinite number of solutions. Here, we will learn about two linear equations in 2 variables. (Both equations having to same variable i.e., x, y)

**Simultaneous linear equations:**

Two linear equations in two variables taken together are called simultaneous linear equations.

The solution of system of simultaneous linear equation is the ordered pair (x, y) which satisfies both the linear equations.

**Necessary steps for forming and solving simultaneous linear equations**

Let us take a mathematical problem to indicate the necessary steps for forming simultaneous equations:

*In a stationery shop, cost of 3 pencil cutters exceeds the price of 2 pens by $2. Also, total price of 7 pencil cutters and 3 pens is $43.*

Follow the steps of instruction along with the method of solution.

**Step I:** Indentify the unknown variables; assume one of them as * x* and the other as

Here two unknown quantities (variables) are:

Price of each pencil cutter = $x

Price of each pen = $y

**Step II:** Identify the relation between the unknown quantities.

Price of 3 pencil cutter =$3x

Price of 2 pens = $2y

Therefore, first condition gives: 3x – 2y = 2

**Step III:** Express the conditions of the problem in terms of * x* and

Again price of 7 pencil cutters = $7x

Price of 3 pens = $3y

Therefore, second condition gives: 7x + 3y = 43

Simultaneous equations formed from the problems:

3x – 2y = 2 ----------- (i)

7x + 3y = 43 ----------- (ii)

**For examples:**

(i) x + y = 12 and x – y = 2 are two linear equation (simultaneous equations). If we take x = 7 and y = 5, then the two equations are satisfied, so we say (7, 5) is the solution of the given simultaneous linear equations.

(ii) Show that x = 2 and y = 1 is the solution of the system of linear equation x + y = 3and 2x + 3y = 7

Put x = 2 and y = 1 in the equation x + y = 3

L.H.S. = x + y = 2 + 1 = 3, which is equal to R.H.S.

In **2ⁿᵈ equation**, 2x + 3y = 7, put x = 2 and y = 1 in L.H.S.

L.H.S. = 2x + 3y = 2 × 2 + 3 × 1 = 4 + 3 = 7, which is equal to R.H.S.

Thus, x = 2 and y = 1 is the solution of the given system of equations.

Worked-out problems on solving simultaneous linear equations:

**1.** x + y = 7 ………… (i)

3x - 2y = 11 ………… (ii)

**Solution:**

The given equations are:

x + y = 7 ………… (i)

3x - 2y = 11 ………… (ii)

From (i) we get y = 7 – x

Now, substituting the value of y in equation (ii), we get;

3x - 2 (7 - x) = 11

or, 3x - 14 + 2x = 11

or, 3x + 2x - 14 = 11

or, 5x - 14 = 11

or, 5x -14 + 14 = 11 + 14 [add 14 in both the sides]

or, 5x = 11 + 14

or, 5x = 25

or, 5x/5 = 25/5 [divide by 5 in both the sides]

or, x = 5

Substituting the value of x in equation (i), we get;

x + y = 7

Put the value of x = 5

or, 5 + y = 7

or, 5 – 5 + y = 7 – 5

or, y = 7 – 5

or, y = 2

Therefore, (5, 2) is the solution of the system of equation **x + y = 7** and **3x – 2y = 11**

**2.** Solve the system of equation 2x – 3y = 1 and 3x – 4y = 1.

**Solution:**

The given equations are:

2x – 3y = 1 ………… (i)

3x – 4y = 1 ………… (ii)

From equation (i), we get;

2x = 1 + 3y

or, x = ¹/₂(1 + 3y)

Substituting the value of x in equation (ii), we get;

or, 3 × ¹/₂(1 + 3y) – 4y = 1

or, ³/₂ + ⁹/₂y - 4y = 1

or, (9y – 8y)/2 = 1 - ³/₂

or, ¹/₂y = (2 – 3)/2

or, ¹/₂y = \(\frac{-1}{2}\)

or, y = \(\frac{-1}{2}\) × \(\frac{2}{1}\)

or, y = -1

Substituting the value of y in equation (i)

2x – 3 × (-1) = 1

or, 2x + 3 = 1

or, 2x = 1 - 3
or, 2x = -2

or, x = -2/2

or, x = -1

Therefore, x = -1 and y = -1 is the solution of the system of equation

**2x – 3y = 1** and **3x – 4y = 1**.

● **Simultaneous Linear Equations**

**Solvability of Linear Simultaneous Equations**

**Word Problems on Simultaneous Linear Equations**

**Word Problems on Simultaneous Linear Equations**

**Practice Test on Word Problems Involving Simultaneous Linear Equations**

● **Simultaneous Linear Equations - Worksheets**

**Worksheet on Simultaneous Linear Equations**

**Worksheet on Problems on Simultaneous Linear Equations**

**8th Grade Math Practice****From Simultaneous Linear Equations to HOME PAGE**

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