* Linear Equations in two variables
- Meaning: Equation with degree one and in terms of two variables
e.g : 2x + 4y -6 = 0
Here degree = 1 [Highest power of variable] in terms of ‘x’ and ‘y’
• Standard form (General form)
ax + by + c = 0
‘a’ and ‘b’ ≠ 0
• Single linear equation has infinite solutions
For e.g : x + y = 5
Here if x=1 y =4
x= -1 y=6
x=0 y=5
This way we can have infinite values of (x, y) which can satisfy equation.
• Simultaneous linear equations
When we think about two linear equations in two variables at the same time, they are called simultaneous equations.
Types of solutions
Let a1x + b1y +c1 =0
a2x +b2y +c2 =0 be linear equations in two variables
Take ratios
a1/a2, b2/b2, c1/c2
From the table above, you can observe that if the lines represented by the equation
a1x + b1y + c1 =0
and
a2x+ b2y + c2 =0
Pair Of Linear Equations In Two Variables
Are (i) intersecting, then a1/a2 ≠ b1/b2.
(ii) coincident, then a1/a2 =b1/b2=c1/c2.
(iii) parallel, then a1/a2 =b1/b2 ≠ c1/c2.
Different Methods to solve Linear equations in two variables
1) Elimination Method
E.g. 1: 5x + 3y-35 =0
2x+4y -28 =0
Step 1:
Write equations in a form
ax + by =c
i.e. 5x+3y =35 — — (I)
2x+4y=28 — — (II)
Step 2:
Check if the coefficient for x or y variables is the same if not make the coefficient of any one variable the same.
In this e.g: if we consider ‘x’ variable To make coefficient same
We need to multiply
eqᶯ (I) by 2 and
eqᶯ (II) by 5
5x × 2 + 3y × 2 = 35 × 2
2x × 5 + 4y × 5 = 28 × 5
We get,
10x + 6y =70 — — — (I)
10x + 20Y =140 — — — (II)
Step 3 :
After making the coefficient same check for the signs of the coefficient which we have made the same.
If signs are same
Subtract the equations
If signs are different
Add the equations
In our case signs of 10’x’ are same in (III) and (IV) equations.
Subtract equation (III) From (IV)
10x + 20 y =140
10x + 6y =70
(-) (-) (-)
— — — — — — —
14y = 70
Y =70/14
y = 5
Step 4: — Once we get value of one variable we can substitute that in any one equation to get value of other variable.
Substitute y = 5 in equation (I)
5x + 3 y =35
5x + 3x (5) = 35
5x + 15 = 35
5x = 35–15
5x = 20
x = 20/5
x = 4
Solution : x=4, y=5
Eg2: 3x + 2y = 29
5x — y = 18
[In this case they are already in form ax + by = C]
So
3x + 2y = 29 — — — — — (1)
5x — y = 18 — — — — — — — (2)
[If we see here ‘x’ coefficients and ‘y’ coefficients, in this case it's easy to make ‘y’ coefficient same by multiplying Eqn(2) By ‘2’]
Multiply equation (2) By ‘2’
5x × 2 -y × 2 = 18 × 2
10x -2y = 36 — — — — — -(3)
[Now we can see coeff of ‘y’ in equation 1 and 3 are same but have different signs]
Add equations 1 and 3
3x + 2y = 29
10x — 2y = 36
— — — — — — — — — — — —
13x =65
𝑥 = 65/13
x= 5
Substitute x = 5 n eqn (1)
3x + 2y =29
3 × 5 + 2y = 29
2y = 29–15
2y = 14
Y = 14/2
Y = 7
Solution : -x = 5 and y = 7
Practice Sums –
1) 2x — 3y = 4 ; x — 3y = -4
2) 3x — 4y = -7 ; 5x — 2y = 0
3) x + y = 6 ; x — y =4
