**Explanation of Algebraic Methods of solving Pair of Linear Equations in two Variables**

__Substitution Method of solving__

__Pair of linear equations in two variables__

**
**

**Let Given Pair of linear equations in two variables**

**Putting eqn. (3) in eqn. (2) we get**

**Putting y = 4 in eqn. (3) we get **

**x = -1 and y = 4 is the required solution.**

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__Elimination Method of solving__

__Pair of linear equations in two variables__

**Eqn. (1) x 2 – eqn. (2) x 1 we get **

**Putting y = 4 in equation (1)**

**x = -1 and y = 4 is the required solution.**

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__Cross Multiplication Method of solving(Deleted)__

__Pair of linear equations in two variables__

**There are two cases to explain this method**

**Case I : Let general pair of linear equations in two variables :**

**\[a_{1}x+b_{1}y=c_{1}\]\[a_{2}x+b_{2}y=c_{2}\]**

**If constant terms on the right hand side then place the coefficients as shown below**

**Case II : If constant terms are on right hand side then we must follow the following steps**

**General pair of linear equations in two variables**

**\[a_{1}x+b_{1}y+c_{1}=0\]\[a_{2}x+b_{2}y+c_{2}=0\]**

**Explanation of cross multiplication method by taking an example**

**Let given a pair of equations in two variables**

**Exercise 3.7**

**Q1 ****The ages of
two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old
as Ani and Biju is twice as old as his
sister cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of
Ani and Biju.**

**Sol. : Let age of Ani = x**

** Let age of Biju = y**

**According to the question**

** x - y = 3 .....................(1)**

**Age of Dharam = 2x**

** Age of cathy = y/2**

**A.T.Q 2x - y/2 = 30**

**Multiplying this equation by 2 we get**

** 4x - y = 60 ....................(2)**

**Solving equation (1) and eqn(2) by elimination method we get**

** x = 19 years, y=16 years.**

**Q2. One
says, “Give me a hundred, friend! I shall then become twice as rich as you”.
The other replies, “If you give me ten, I shall be six times as rich as you”.
Tell me what is the amount of their capital?**

**Sol. : Let first friend have = Rs x**

** Second friend have = Rs y**

**A.T.Q. **

** x + 100 = 2(y - 100)**

** x + 100 = 2y - 200**

** x - 2y = - 200 - 100**

** x - 2y = - 300 ...............(1)**

**Again ATQ**

** 6(x - 10) = y + 10**

** 6x - 60 = y + 10**

** 6x - y = 10 + 60**

** 6x - y = 70 ......................(2)**

**Solving equation (1) and (2) by elimination method we get**

** x = 40 and y = 170**

**Q3. A train
covered a certain distance at a uniform speed. If the train would have been 10
km/h faster, it would have taken 2 hours less than the scheduled time. And if
the train were slower by 10 km/h, it would have taken 3 hours more than the
scheduled time. Find the distance covered by the train.**

**Sol. : Let speed of train = x km/h**

** Time taken by train = y Km/h**

** Distance = Speed x Time**

** = xy**

**Case 1 **

** New speed = (x + 10) km/h**

** New time taken = y - 2) hours**

** Distance = Speed x Time**

** = (x + 10) x (y - 2)**

** = xy - 2x + 10y - 20**

**A.T.O. **

** **~~xy~~ - 2x + 10y - 20 = ~~xy~~

** -2x + 10y = 20 .......................(1)**

**Case II**

** New speed = (x - 10) km/h**

** New time taken = (y + 3) hours**

**Distance = (x - 10) x (y + 3)**

** = xy + 3x - 10y - 30**

**A.T.Q.**

** **~~xy~~ + 3x - 10y - 30 = ~~xy~~

** 3x - 10y - 30 = 0**

** 3x - 10y = 30 .....................(2)**

**By elimination method solve equation (1) and (2) we get**

** x = 50, y = 12**

**Distance = xy = 50 x 12 = 600km**

**Q4. The
students of a class are made to stand in rows. If 3 students are extra in each
row, there would be 1 row less. If 3 students are less in each row, there would
be 2 rows more. Find the number of students in the class.**

**Sol. Let number of rows = x**

** Number of students in each row = y**

** Total students in the class = xy**

**A.T.Q.**

** (x + 3)(y - 1) = xy**

** **~~xy~~ - x + 3y - 3 = ~~xy~~

** - x + 3y - 3 = 0**

** - x + 3y = 3 .................(1)**

**Again ATQ**

** (x - 3)(y + 2) = xy**

** **~~xy~~ + 2x - 3y - 6 = ~~xy~~

** 2x - 3y = 6 ....................(2)**

**Solving equations (1) and (2) by elimination method we get**

**x = 9, y = 4**

**Number of students in the class = 9 x 4 = 36**

**Q7. Solve the following equations**

**i) Solve the following ****By cross multiplication method**** **

**Solution**

**ii) Solve the following equations ****By Cross multiplication method**

**ax + by = c**

**bx + ay = 1+c**

**iii) Solve : \[\frac{x}{a}-\frac{y}{b}=0 ........(1)\]\[ax+by=a^{2}+b^{2} .......(2)\]**

From equation (1)

**Putting equation (3) in equation (2) we get**

**Putting y = b in equation (3) we get**

**Q7 iv) ****(a - b)x + (a
+ b)y = a**^{2} - 2ab - b^{2} …..(1)

** (a + b)(x +
y) = a**^{2} + b^{2}

** (a + b)x + (a
+ b)y = a**^{2} + b^{2} …….(2)

**Eqn. (1) –
Eqn.(2)**

** - **~~2b~~x =
- ~~2b~~(a – b) ⇒ **x = a – b**

**Putting x =
a - b in equation (2)**

**(a + b)(a - b)
+ (a + b)y = a**^{2} + b^{2}

^{}

** **** **~~a~~^{2} - b^{2} + (a + b)y = ~~a~~^{2} + b^{2}

**Q7(v) **** 152x – 378y = -74 …….(1)**

** -378x +
152y = - 604 ……(2)**

**Adding
eqn.(1) and eqn.(2)**

**Dividing by
-226 we get**

** x + y = 3 …………..(3)**

**Subtracting
eqn.(1) and eqn.(2) we get **

**Dividing by
530 we get**

** x – y =1 ……………(4)**

**By
elimination method solving eqn. (3) and (4) we get **

** x = 2, y = 1**

**Q 8)**

**ABCD
is a cyclic **

**quadrilateral,
find the measure of angles of the cyclic quadrilateral ?**

**Sol. Since opposite angles of a cyclic quadrilateral are supplementary.**

**Therefore**

** ∠A + ****∠C = ****180**^{o}

** 4y + 20 - 4x = 180**

** - 4x + 4y = ****180 ****- 20**

** - 4x + 4y = ****160**

**Dividing by 4 we get**

** -x + y = 4****0****..........(1)**

** ∠B + ****∠D = ****180**^{o}

** 3y - 5 - 7x + 5 = ****180**

** -7x + 3y = ****180**** ..................(2)**

**Solving these equations by elimination method:-**

**Eqn.(1) x 3 - eqn. (2) x 1 we get**

**x = -60/4 ⇒ x = -15**

**Putting x = -15 in eqn. (1) we get**

** -(-15) + y = 40**

** y = 40 - 15**

** y = 25**

** x = -15, y = 25**

**Now we find the angles of quadrilateral**

** ∠A = 4y + 20 = 4 x 25 + 20 = 12****0**^{o}

** ∠B = 3y - 5 = 3 x 25 - 5 = 7****0**^{o}

** ∠C = - 4x = -4 x -15 = 6****0**^{o}

** ∠D = -7x + 5 = -7 x - 15 + 5 = **** 105 + 5 = 11****0**^{o}

**THANKS FOR YOUR VISIT**

**PLEASE COMMENT BELOW**

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