Pair of Linear Equations in Two Variables
Mathematics Class Tenth
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:
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,
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,
`=>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
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
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