Pair of Linear Equations in Two Variables
Mathematics Class Tenth
NCERT Exercise 3.4
Elimination Method
Solving a pair of linear equations by eliminating one variable is called Elimination Method. Elimination method is more convenient than the Substitution method.
Step to solve a pair of linear equations by Elimination Method
Step: 1. First multiply both the equations by some suitable non–zero constants to make the coefficients of one variable (either `x` or `y`) numerically equal.
Step2: Then add or subtract one equation from the other so that one variable gets eliminated. If an equation in one variable is obtained go to Step: 3.
If in Step 2, we obtain a true statement involving no variable, then the original pair of equations has infinitely many solutions.
If in Step 2, we obtain a false statement involving no variable, then the original pair of equations has no solution, i.e., it is inconsistent.
Step: 3: Solve the equation in one variable (`x` or `y`) so obtained to get its value.
Step: 4: Substitute this value of `x` (or `y`) in either of the original equations to get the value of the other variable.
Example:
Let, a pair of linear equations in two variables
`2x+3y = 35` ---------(i)
`3x+2y = 40` --------(ii)
Step: 1 Multiply equation (i) by 3
i.e. `3(2x+3y =35)`
`=6x+9y = 105` ---------(iii)
And multiply equation (ii) by 2
i.e. `2(3x+2y = 40)`
`=> 6x+4y = 80` -----------(iv)
Step: 2. Now, subtract equation (iv) from equation (iii)
`(6x+9y) - (6x+4y) = 105-80`
Step: 3.
`=> \cancel(6x)+9y - \cancel(6x)-4y = 25`
`=> 9y - 4y = 25`
`=> 5y = 25`
`:. y = 25/5 = 5`
Step: 4:
Now, substitute the value of `y` in equation (i)
`2x+3y = 35`
`=> 2x+3xx5 = 35`
`=> 2x +15 = 35`
`=> 2x = 35-15 = 20`
`:. x = 20/2 = 10`
Thus, `x = 10` and `y=5` Answer
NCERT Exercise 3.4
Question: 1. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) `x+y = 5` and
`2x-3y = 4`
Solution: Given,
`x+y = 5` ---------(i)
`2x-3y = 4` --------(ii)
After multiplying equation (i) by 3, we get
`3(x+y = 5)`
`=>3x+3y = 15` -------(iii)
Now, after adding equation (ii) and equation (iii), we get
`2x-\cancel(3y) + 3x+\cancel(3y) = 4 + 15`
`=> 2x+3x = 19`
`=> 5x = 19`
`=> x = 19/5`
Now, after substituting the value of `x` in equation (i), we get
`19/5+y = 5`
`=> y = 5-19/5`
`=> y = (25-19)/5 = 6/5`
Thus, `x= 19/5` and `y = 6/5` Answer
(ii) `3x+4y = 10`;
`2x-2y = 2`
Solution:
Given, `3x+4y = 10` ---------- (i)
`2x-2y = 2` ---------- (ii)
After multiplying equation (ii) by 2, we get
`2(2x-2y = 2)`
`=> 4x-4y = 4` ----------(iii)
After adding equation (i) and equation (iii), we get
`(3x+4y)+(4x-4y) = 10+4`
`=> 3x+\cancel(4y)+4x-\cancel(4y) = 14`
`=>7x = 14`
`:. x = 14/7 = 2`
Now, after substituting the value of `x` in equation (ii), we get
`2xx2 - 2y = 2`
`=> 4-2y = 2`
`=> -2y = 2-4`
`=> -2y = -4`
`=> y = (-4)/(-2) = 1`
Thus, `x = 2` and `y = 1` Answer
(iii) `3x-5y-4=0`;
`9x=2y+7`
Solution:
Given, `3x-5y-4=0` ----------- (i)
`9x=2y+7`
`=> 9x - 2y - 7 = 0` ------------(ii)
After multiplying equation (i) by 3, we get
`3(3x-5y-4=0)`
`=>9x-15y - 12 =0` ------------(iii)
Now, after subtracting equation (ii) from equation (iii), we get
`(9x-15y-12)-(9x-2y-7) = 0 - 0`
`=>\cancel(9x)-15y-12-\cancel(9x)+2y+7 =0`
`=> -13y - 5 = 0`
`=> -13y = 5`
`=> y = -5/13`
Now, after substituting the value of `y` in equation (i) we get
`3x - 5(-5/13) -4=0`
`=> 3x + 25/13 = 4`
`=> 3x = 4-25/13`
`=> 3x = (52-25)/13`
`=> 3x = 27/13`
`=> x =( \cancel(27)9)/13 xx 1/\cancel(3)`
`=> x = 9/13`
Thus, ` x = 9/13` and `y = -5/13` Answer
(iv) `x/2+(2y)/3=-1`;
`x - y/3 = 3`
Solution:
Given, `x/2+(2y)/3=-1` --------(i)
`x - y/3 = 3` --------------(ii)
After multiplying equation (ii) by 2, we get
`2(x-y/3 =3)`
`=> 2x - (2y)/3=6` ----------(iii)
Now, after adding equation (i) and (ii), we get
`x/2 + \cancel((2y)/3) + 2x - \cancel((2y)/3) = -1+6`
`=> x/2 +2x = 5`
`=> (x+4x)/2 = 5`
After cross multiplication, we get
`x+4x = 5xx2 =10`
`=> 5x = 10`
`=> x = 10/5 = 2`
Now, after substituting the value of `x` in equation (ii), we get
`2 - y/3 = 3`
`=> - y/3 = 3-2 = 1`
After cross multiplication
`=> - y = 3`
`=> y = -3`
Thus, `x=2` and `y = -3` Answer
Question: 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. If becomes `1/2` if we only add 1 to the denominator. What is the fraction?
Solution:
Let, fraction is `x/y`
Thus, according question, after adding 1 to numerator and subtracting 1 from numerator
`(x+1)/(y-1) = 1`
After cross multiplication, we get
`x+1 = y-1`
`=> x = y -1 - 1`
`=> x-y = -2` -----------(i)
And according to question, after adding 1 to the denominator,
`x/(y+1) = 1/2`
After cross multiplication,
`=> 2x = y+1`
`=> 2x - y = 1` --------(ii)
Now, after subtracting equation (ii) from equation (i), we get
`(x-y) - (2x-y) = -2 ? 1`
`=> x- \cancel(y) -2x+\cancel(y) = -3`
`=> -x = -3`
`=> x = 3`
After substituting the value of `x` in equation (ii), we get
`2(3) - y = 1`
`=>6- y = 1`
`=> - y = 1-6 =-5`
`=> y = 5`
Thus, fraction is `3/5` Answer
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution:
Let present age of Nuri `=n`
And present age of Sonu `=s`
Now, Five years ago
Age of Nuri `=n-5`
And age of Sonu `=s-5`
Thus, according to question,
`n-5 = 3(s-5)`
`=> n - 5 = 3s-15`
`=> n -3x = -15+5`
`=> n-3s = -10` --------(i)
Ten years later,
Age of Nuri `= n +10`
And age of Sonu ` = s+10`
Now, according to question,
`n+10 = 2(s+10)`
`=> n+10 = 2s+20`
`=> n - 2s = 20-10`
`=> n -2s = 10` --------(ii)
Now, by subtracting equation (ii) from equation (i), we get
`(n-3s)-(n-2s) = -10-10`
`=>\cancel(n)-3s-\cancel(n)+2s = -20`
`=>-s = -20`
`=> s = 20`
Now, after substituting the value of `s` in equation (ii), we get
`n - 2xx20 = 10`
`=> n -40 = 10`
`=> n = 10+40 = 50`
Thus, present age of Nuri = 50 years and present age of Sonu = 20 years. Answer
(iii) The sum of the digits of a two digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let the unit digit of the number `=x`
And the tens digits of the number `=y`
Thus, number `=10y +x`
Now, according to question,
`x+y = 9` ------------(i)
Reverse of the number `= 10x+y`
Thus, according to question,
`9(10y+x) = 2(10x+y)`
`=> 90y + 9x = 20x + 2y`
`=> 90y - 2y +9x-20x = 0`
`=>88y - 11x=0`
`=>11(8y-x)=0`
`=>8y - x =0` -----------(ii)
Now, after adding equation (i) and equation (ii), we get
`(x+y)+(8y-x)=9+0`
`=>\cancel(x)+y +8y-\cancel(x) = 9`
`=>9y = 9`
`=> y = 9/9=1`
Now, after substituting the value of `y` in equation (i), we get
`x +1 = 9`
`=> x = 9-1 = 8`
Thus, number is `10y +x = 10xx1+8=18`
Thus, number `=18` Answer.
(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
Solution:
Let, number of Rs 50 notes `=x`
And the number of Rs 100 notes `=y`
Thus, according to question,
`x+y = 25` -----------(i)
And `50x+100y = 2000`
`=> 50(x+2y)= 2000`
`=>x+2y = 2000/50`
`=>x+2y = 40` ---------(ii)
Now, after subtracting equation (i) from equation (ii), we get
`(x+2y) - (x+y) = 40 - 25`
`=> \cancel(x)+2y-\cancel(x)-y=15`
`=>y = 15`
After substituting the value of `y` in equation (i), we get
`x+15=25`
`=> x = 25-15 = 10`
Thus, number of Rs 50 notes = 10 and number of Rs 100 notes = 15. Answer
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
Let the fixed charge for the first three days = Rs `x`
And additional charge for each day = Rs `y`
Thus, according to question,
In the case of charge paid by Saritha for 7 days (3 fixed days + additional 4 days)
`x + 4 y = 27` ---------(i)
And in the case of charge paid by Susy for 5 days (3 fixed days + additional 2 days)
`x+2y = 21` -------------(ii)
Now, by subtracting equation (ii) from equation (i), we get
`(x+4y)-(x+2y)=27-21`
`=> \cancel(x)+4y-\cancel(x)-2y = 6`
`=> 2 y = 6`
`=> y = 6/2 = 3`
After subtracting value of `y` in equation (ii), we get
`x+2xx3 = 21`
`=> x +6=21`
`=> x = 21-6 = 15`
Thus, fixed charge for three days = Rs `15` and additional charge for each day = Rs `3` Answer
Reference: