Pair of Linear Equations in Two Variables

Mathematics Class Tenth

10th-Math-home


NCERT Exercise 3.5

Cross Multiplication Method to Solve a Pair of Linear Equation

Although all methods to solve a Pair of Linear Equation are good and have equal importance but Sometimes Cross Multiplication method is proved more convenient than other.

Let a Pair of Linear Equation

`a_1x+b_1y+c_1 = 0` and

`a_2+b_2y + c_2 = 0`

Therefore,

`x = (b_1c_2 - b_2c_1)/(a_1b_2-a_2b_1)`, provided `a_1b_2 - a_2b_1 !=0`

And, `y = (c_1a_2-c_2a_1)/(a_1b_2 ? a_2b_1)`

Now two cases arise:

Case : 1– `a_1b_2 - a_2b_1 !=0`.

In this case `a_1/a_2!=b_1/b_2`.

Then the pair of linear equations has unique solution.

Case : 2– `a_1b_2 - a_2b_1 =0`.

If write `a_1/a_2 = b_1/b_2 = k`,

Then `a_1 = ka_2`, and `b_1 = kb_2`.

Therefore, pair of linear equations has infinitely many solutions.

To use Cross Multiplication write the pair of equation in the following form:

10 math pair of linear equation in two variables 1 ncert exercise 3.5

Thus,

`x/(b_1c_2-b_2c_1) = y/(c_1a_2-c_2a_1)` `=1/(a_1b_2-a_2b_1)`

Thus,

`x/(b_1c_2-b_2c_1) =1/(a_1b_2-a_2b_1)`

`y/(c_1a_2-c_2a_1)=1/(a_1b_2-a_2b_1)`

Here, if

`a_1/a_2!=b_1/b_2`, then pair of linear equations has unique solution.

And if `a_1/a_2 = b_1/b_2 = c_1/c_2`, the pair of linear equations has infinitely many solutions.

And if `a_1/b_1 = b_1/b_2!=c_1/c_2`, then pair of linear equations has no solution.

NCERT Exercise 3.5

Question: 1. Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(i) `x-3y-3=0`;

`3x-9y-2=0`

Solution:

Given, `x-3y-3=0`;

`3x-9y-2=0`

By comparing given equations with `a_1x+b_1y+c_1=0` and `a_2+b_2y+c_2=0`, we get

`a_1 = 1, b_1 = -3, c_1 = -3` and

`a_2 = 3, b_2 = -9, c_2 = -2`

Thus,

`a_1/a_2 = 1/3`

And, `b_1/b_2 = (-3)/(-9) = 1/3`

And, `c_1/c_2 = (-3)/(-2) = 3/2`

Here, since `a_1/a_2 = b_1/b_2 !=c_1/c_2` thus, given pair of linear equations has no solution.

(ii) `2x+y = 5`

`3x+2y = 8`

Solution:

Given, `2x+y = 5`

`3x+2y = 8`

By comparing given equations with `a_1x+b_1y+c_1=0` and `a_2+b_2y+c_2=0`, we get

`a_1 = 2, b_1 = 1, c_1 = -5`

And, `a_2 = 3, b_2 = 2, c_2 = -8`

Thus,

`a_1/a_2 = 2/3`

And `b_1/b_2 = 1/2`

And, `c_1/c_2 = (-5)/(-8) = 5/8`

Here, since, `a_1/a_2 !=b_1/b_2`

Thus, given pair of linear equation has unique solution.

By applying,

10 math pair of linear equation in two variables 1 ncert exercise 3.5

10 math pair of linear equation in two variables 2 ncert exercise 3.5

We get,

`x/(b_1c_2-b_2c_1) = y/(c_1a_2-c_2a_1)` `=1/(a_1b_2-a_2b_1)`

`x/(1(-8)-2(-5)) = y/((-5)3-(-8)2)` `=1/(2xx2-3xx1)`

`=>x/(-8+10) = y/(-15+16)=1/(4-3)`

`=>x/2=y/1=1/1`

Now, when, `x/2 = 1/1`

`:. x = 2`

And, when `y/1=1/1`

`:. y =1`

Thus, `x = 2` and `y=1` Answer

(iii) `3x-5y=20`;

`6x-10y=40`

Solution:

Given, `3x-5y=20`;

`=>3x-5y-20=0` ----------(i)

`6x-10y=40`

`=>6x-10y-40=0` ------------(ii)

By comparing given pair of linear equations with `a_1x+b_1y+c_1=0` and `a_2+b_2y+c_2=0`, we get

`a_1 = 3, b_1 =-5, c_1 = -20`

And, `a_2 = 6, b_2 = -10, c_2 = -40`

Thus, `a_1/a_2 = 3/6 = 1/2`

And, `b_1/b_2 = (-5)/(-10) = 1/2`

And, `c_1/c_2 = (-20)/(-40) = 1/2`

Here, since, `a_1/a_2=b_1/b_2=c_1/c_2`

Thus, given pair of linear equations has infinitely many solutions.

(iv) `x-3y-7=0`;

`3x-3y-15=0`

Solution:

Given, `x-3y-7=0` ---------(i)

`3x-3y-15=0` ---------(ii)

By comparing given pair of linear equations with `a_1x+b_1y+c_1=0` and `a_2+b_2y+c_2=0`, we get

`a_1 = 1, b_1 = -3, c_1 = -7`

And, `a_2 = 3, b_2 = -3, c_2 =-15`

Now,

`a_1/a_2 = 1/3`

And, `b_1/b_2 = (-3)/(-3)=1`

And, `c_1/c_2 = (-7)/(-15) = 7/15`

Here, since, `a_1/a_2 !=b_1/b_2`, thus given pair of linear equations has unique solution.

Now, we know that,

10 math pair of linear equation in two variables 1 ncert exercise 3.5

`=>x/(b_1c_2-b_2c_1) = y/(c_1a_2-c_2a_1)` `=1/(a_1b_2-a_2b_1)`

Thus, by applying cross multiplication rule

10 math pair of linear equation in two variables 3 ncert exercise 3.5

Thus,

`x/((-3)(-15)-(-3)(-7))` ` = y/((-7)3 - (-15)1)` `=1/(1(-3)-3(-3))`

`=>x/(45-(21))=y/(-21-(-15))` `=1/(-3+9)`

`=>x/24 = y/(-21+15)=1/6`

`=>x/24=y/(-6)=1/6`

Now, when, `x/24 = 1/6`

Therefore, `6x = 24`

`=> x = 24/6 =4`

And, when, `y/(-6) = 1/6`

`=> 6y = -6`

`=> y = (-6)/6 = -1`

Thus, `x = 4` and `y=-1` Answer

Question: 2. (i) For which values of `a` and `b` does the following pair of linear equations have an infinite number of solutions?

`2x+3y = 7`

`(a-b)x+(a+b)y` `=3a+b-2`

Solution:

Given, `2x+3y = 7`

`(a-b)x+(a+b)y` `=3a+b-2`

By comparing given pair of linear equations with `a_1x+b_1y+c_1=0` and `a_2+b_2y+c_2=0`, we get

`a_1 = 2, b_1 = 3, c_1 = -7`

And, `a_2 = a-b, b_2 = a+b`, `c_2 = -(3a+b-2)`

Now, for infinite number of solutions a pair of linear equations must have

`a_1/a_2=b_1/b_2=c_1/c_2`

Thus, from given pair of linear equations after substituting the values of `a_1, a_2, b_1, b_1`, and `c_1, c_2`, we get

`2/(a-b) = 3/(a+b)=(-7)/(-(3a+b-2))`

Now,` 2/(a-b) = 7/(3a+b-2)`

After cross multiplication, we get

`2(3a+b-2)=7(a-b)`

`=> 6a+2b-4 = 7a-7b`

`=>6a-7a+2b+7b-4=0`

`=>-a+9b -4=0`

`=>-(a-9b+4)=0`

`=>a-9b+4=0` ----------(i)

And, now,

`2/(a-b) = 3/(a+b)`

After cross multiplication, we get

`2(a+b) = 3(a-b)`

`=>2a+2b=3a-3b`

`=>2a-3a+2b+3b=0`

`=>-a+5b=0` ----------(ii)

After adding equation (i) and (ii), we get

`(a-9b+4)+(-a+5b)=0`

`=>\cancel(a)-9b+4-\cancel(a)+5b=0`

`=>-4b+4=0`

`=>4b = 4`

`:. b = 4/4 = 1`

Now, by substituting the value of `b` in equation (ii), we get

`-a+5xx1=0`

`=>-a+5=0`

`=>a = 5`

Thus, for the values of `a=5` and `b=1` given pair of linear equations have infinite number of solutions. Answer

(ii) For which value of `k` will the following pair of linear equations have no solution?

`3x+y=1`

`(2k-1)x+(k-1)y=2k+1`

Solution:

Given, `3x+y=1`

`(2k-1)x+(k-1)y=2k+1`

By comparing given pair of linear equations with `a_1x+b_1y+c_1=0` and `a_2+b_2y+c_2=0`, we get

`a_1 = 3, b_1 = 1, c_1 = -1`

And, `a_2 = 2k-1, b_2 = k-1, c_1 = 2k+1`

Here, `a_1/a_2 =3/(2k-1)`

And, `b_1/b_2 = 1/(k-1)`

And, `c_1/c_2 = (-1)/(2k+1)`

Since, pair of linear equations have no solution only if,

`a_1/a_2=b_1/b_2!=c_1/c_2`

Thus, `3/(2k-1) = 1/(k-1)`

After cross multiplication, we get

`3(k-1) = 1(2k-1)`

`=> 3k - 3 = 2k-1`

`=>3k-2k = -1+3`

`=>k = 2`

Thus, for the value of `k=2` given pair of equations has no solution. Answer

MCQs Test

Back to 10th-Math-home



Reference: