Pair of Linear Equations in Two Variables
Mathematics Class Tenth
NCERT Exercise 3.5-(2)
NCERT Exercise 3.5-Part-2
Question: 3. Solve the following pair of linear equations by the substitution and cross multiplication methods:
`8x+5y=9`
`3x+2y=4`
Solution:
Given, `8x+5y=9` ---------(i)
`3x+2y=4` --------- (ii)
Substitution Method
From equation (i)
`8x = 9-5y`
`=>x =(9-5y)/8` -------- (iii)
Now, by substituting the value of `x` from equation (iii) into equation (ii), we get
`3xx(9-5y)/8+2y = 4`
`=>((27-15y)+16y)/8 = 4`
`=>(27+y)/8=4`
After cross multiplication, we get
`=> 27+y = 8xx4=32`
`=>y = 32-27=5`
Now, after substituting the value of `y` in equation (i), we get
`8x+5xx5=9`
`=>8x+25=9`
`=>8x=9-25=-16`
`=>x = -16/8=-2`
Thus, `x=-2` and `y=5`
Cross Multiplication Method
Given, `8x+5y=9` ---------(i)
`3x+2y=4` --------- (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=8, b_1 = 5, c_1 = -9`
And `a_2=3, b_2 = 2, c_2 = -4`
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/(5(-4)-2(-9))=y/(-9xx3-(-4xx8))` `=1/(8xx2-3xx5)`
`=>x/(-20+18)=y/(-27+32)=1/(16-15)`
`=>x/(-2)=y/5=1/1`
Now, when, `x/(-2) = 1/1`
`:. x = -2`
And when, `y/5 = 1/1`
`:. y = 5`
Thus, `x=-2` and `y=5` Answer.
Question: 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method.
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution:
Let fixed charge for hostel `=x` Rs
And the charge per day for taking food `=y` Rs
Thus, according to question,
In the case of student A for taking food for 20 days
`x+20y = 1000` -----------(i)
In the case of student B for taking food for 26 days
`x+26y = 1180` -------------(ii)
This pair of linear equation is getting solved by Elimination method
Now after subtracting equation (i) from equation (ii), we get
`x+26y -(x+20y)=1180-1000`
`=>\cancel(x)+26y-\cancel(x)-20y=180`
`=>6y = 180`
`=>y = 180/6`
`=> y = 30`
Now, by substituting the value of `y` in equation (i), we get
`x+20xx30=1000`
`=>x+600=1000`
`=>x=1000-600=400`
Thus, fixed charge = Rs `400` and charge for food per day = Rs `30` Answer
(ii) A fraction becomes `1/3` when 1 is subtracted from the numerator and it becomes `1/4` when 8 is added to its denominator. Find the fraction.
Solution:
Let, fraction `=x/y`
Thus, according to question after subtracting 1 from numerator,
`(x-1)/y=1/3`
After cross multiplication, we get
`3(x-1)=y`
`=>3x-3=y`
`=>3x-y-3=0`
`=>3x-y=3` --------- (i)
And, according to question, after addition of `8` to the denominator of fraction,
`x/(y+8)=1/4`
After cross multiplication, we get
`=>4x=y+8`
`=>4x-y=8` ------------ (ii)
This pair of linear equation is getting solved by Elimination method,
Now by subtracting equation (i) from equation (ii), we get
`(4x-y)-(3x-y)=8-3`
`=>4x-\cancel(y)-3x+\cancel(y)=5`
`=>x =5`
Now, after substituting the value of `x` in equation (ii), we get
`4xx5-y=8`
`=>20-y = 8`
`=>-y = 8-20 = -12`
`=>y = 12`
Thus, fraction is `5/12` Answer
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution:
Let number of right answer `=r`
And number of wrong answer `=w`
Thus, according to question, when Yash gets 40 marks
`3r-w =40` ----------------(i)
And when Yash would have 50 marks
`4r-2w=50`
`=>2(2r-w)=50`
`=>2r-w=50/2`
`=>2r-w=25` -------------(ii)
Substitution Method:
From equation (ii)
`=>2r-w=25`
`=>2r-25=w` ----------- (iii)
Now, by substituting value of `w` in equation (i), we get
`3r-(2r-25)=40`
`=>3r-2r+25=40`
`=>r = 40-25=15`
Now, by substituting the value of `r` in equation (iii), we get
`w=2xx15-25`
`=>w=30-25=5`
Thus, number of right answer = 15
And number of wrong answer = 5
And thus total number of questions = 15+5 = 20 Answer
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution:
Let the speed of car A = a km/hour
And the speed of car B = b km/hour
Thus, when they travel in same direction, relative speed of cars = `a-b\ ` km/hour
And when they travel in opposite direction, relative speed of cars `=a+b\ ` km/hour
Now, we know that, time × speed = distance
Given, distance = 100 km
And time taken when they travel in same direction = 5 hour
And time taken when they travel in opposite direction = 1 hour
Thus, when cars travel in same direction
`5(a-b) = 100`
`=>a-b=100/5`
`=>a-b=20` ------------ (i)
When cars travel in opposite direction
`1(a+b) = 100`
`=>a+b=100` ----------- (ii)
Now, after adding equation (i) and equation (ii), we get
`(a-b)+(a+b) = 20+100`
`=>a-\cancel(b)+a+\cancel(b)=120`
`=>2a = 120`
`=>a=120/2=60`
Now, by substituting the value of `a` in equation (i), we get
`60-b=20`
`=>-b=20-60=-40`
`=>b=40`
Thus, speed of car A = 60 km/h and speed of car B = 40 km/h. Answer
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
Let length of the given rectangle `=x` units
And breadth of the given rectangle `=y` units
Thus, area = length `xx` breadth =`xy` square units
Now, according to question,
When, length is reduced by 5 units and breadth is increased by 3 units,
`(x-5)(y+3) = xy - 9`
`=>xy+3x-5y-15=xy-9`
`=>3x-5y=xy-9-xy+15`
`=>3x-5y=6` ------------- (i)
And according to question when length is increased by 3 units and breadth is increased by 2 units,
`(x+3)(y+2)=xy+67`
`=>xy+2x+3y+6=xy+67`
`=>2x+3y+6=xy+67-xy`
`=>2x+3y=67-6`
`=>2x+3y=61` ----------- (ii)
After multiplying equation (i) by 2, we get
`2(3x-5y=6)`
`=>6x-10y=12` ------------- (iii)
And after multiplying equation (ii) by 3, we get
`3(2x+3y=61)`
`=>6x+9y=183` ----------- (iv)
Now, after subtracting equation (iii) from equation (iv), we get
`(6x+9y)-(6x-10y)=183-12`
`=>\cancel(6x)+9y-\cancel(6x)+10y=171`
`=>19y=171`
`=>y = 171/19 = 9`
Now, by substituting the value of `y` in equation (i) we get
`3x-5xx9=6`
`=>3x-45=6`
`=>3x=6+45=51`
`=>x=51/3=17`
Thus, length of the rectangle `=17` units and breadth `=9` units. Answer
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