Pair of Linear Equations in Two Variables

NCERT Exercise 3.7(optional)

Question: 1. 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 differs by 30 years. Find the ages of Ani and Biju.

Solution:

Let the age of Ani `=a` years

And let the age of Biju `=b` years

As given, Ani's father Dharam is twice as old as Ani,

Therefore, Age of Ani's father Dharam `=2a`

And as given, Biju is twice as old as his sister Cathy,

Therefore, age of Cathy `=b/2`

And given, The ages of Cathy and Dharam differs by 30 years.

Thus, `2a-b/2=30`

`=>(4a-b)2=30`

After cross multiplication

`=>4a-b=60` ----------------- (i)

Again, given the difference of ages ob Biju and Ani = 3 years.

Thus, two cases arise, either Ani is older than Biju or Biju is older than Ani.

Now, Case: I: If Ani is 3 years older than Biju

Then, `a-b = 3` -------------- (ii)

Thus, after subtracting equation (ii) from equation (i), we get

`=>a = 57/3=19`

After substituting the value of `a` in equation (ii), we get

`19-b=3`

`=>b = 19-3=16`

Thus, age of Ani = 19 years and Age of Biju = 16 years.

And, Case: II. When Biju is 3 years older than Ani,

Therefore, `b-a = 3`

`=>-a+b=3` ------------ (iii)

Now, by adding equation (i) and (iii), we get

`:. a =63/3 = 21`

Now, by substituting the value of `a` in equation (iii), we get

`-21 +b = 3`

`=>b = 3+21=24`

Thus, age of Ani = 21 years and age of Biju = 24 years.

Thus, When Ani is older than Biju, then Ani = 19 years and Biju = 16 years. Answer

And when Biju is older than Ani, then Ani = 21 years and Biju = 24 years. Answer

Question: 2. 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 (respective) capital? [From the Bijaganita of Bhaskara II].

[Hint: `x+100=2(y-100), y+10=6(x-10)`]

Solution:

Let first friend has Rs `x` and Second friend has Rs `y`.

Now, as given hint in question,

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

`=>x+100=2y-200`

`=>x-2y=-200-100`

`=>x-2y=-300` ------------- (i)

`=>x =-300+2y` ---------- (ii)

Again as given hint in question,

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

`=>y+10=6x-60`

`=>6x-y=10+60`

`=>6x-y=70` ------------ (iii)

Now, by substituting the value of `x=-300+2y`, in above equation (iii), we get

`6(-300+2y)-y=70`

`=>-1800+12y-y=70`

`=>11y-1800=70`

`=>11y=70+1800`

`=>11y=1870`

`=>y=1870/11`

`=>y=170`

Now, by substituting the value of `y=170` in equation (iii) [`6x-y=70`], we get

`6x-170=70`

`=>6x=70+170`

`=>6x=240`

`=>x=240/6`

`=>x=40`

Thus, they had Rs 40 and Rs 170 respectively. Answer

Question: 3. 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.

Solution:

Let the initial speed of train `=u\ km//h`

And, time taken by the train to cover the given distance at uniform speed `=t` hours

Let, the given distance `=d` km

Now, we know that, Distance = speed × time

Thus, `d=uxxt` ---------- (i)

Now, according to question

Condition: I. if train runs `10\ km//h` faster,

i.e. speed `= u+10` km/h

Then time `=t-2` hours

Thus, `d = (u+10)(t-2)`

`=>d = ut + 10 t-2u-20`

After substituting the value of `d=ut` from equation (i), we get

`ut = ut+10t-2u-20`

`=>10t-2u-20=ut-ut`

`=>10t-2u=20` --------------- (ii)

Condition: II. If train runs slower by `10\ km//h`

Therefore, speed `=u-10` km/h

And time `=t+3` hours

Therefore, `d=(u-10)(t+3)`

`=>d=ut-10t+3u-30`

After substituting the value of `d=ut` from equation (i), we ge

`ut = ut-10t+3u-30`

`=>-10t+3u-30=ut-ut`

`=>-10t+3u=30` ----------- (iii)

Now, by adding equation (ii) and (iii), we get

`(10t-2u)+(-10t+3u)=20+30`

`=>\cancel(10t)-2u-\cancel(10t)+3u=50`

`=>u=50`

Now, by substituting the value `u=50` in equation (ii) [`10t-2u=20`], we get

`10t-2xx50=20`

`=>10t-100=20`

`=>10t=20+100`

`=>10t=120`

`=>t=120/10`

`=>t=12`

Thus, now after substituting the value of `u=50` and `t=12` in equation (i) [`d=ut`], we get

`d=50xx12`

`=>d = 600`

Thus, distance covered by train = 600 km Answer

Question: 4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Solution:

Let the number of row `=r` and number of students in a row `=n`

Therefore, Total students

= Number of rows × Number of student in a row

`=>s=rxxn` ---------- (i)

Now, as given,

Case: I.

When total number of students is 3 more, then row becomes 1 less

Thus, `s=(r-1)(n+3)`

`=>r\ n= rn+3r-n-3`

[∵ `(s=r\ n)` From equation (i)]

`=>3r-n-3=rn-rn`

`=>3r-n-3=0`

`=>3r-n=3` ------------ (ii)

Now, as per question,

Case; 2.

When 3 students are less the row would be 2 more

Thus, Total number of students `=(r+2)(n-3)`

`=>s=rn-3r+2n-6`

`=>rn=rn-3r+2n-6`

[∵ `(s=r\ n)` From equation (i)]

`=>-3r+2n-6=rn-rn`

`=>-3r+2n=6` ----------- (iii)

Now, by adding equation (ii) and (iii), we get

`(3r-n)+(-3r+2n)=3+6`

`=>\cancel(3r)-n-\cancel(3r)+2n=9`

`=>-n+2n=9`

`=>n=9`

Now, by substituting the value `n=9` in equation (ii) [`3r-n=3`], we get

`3r-9=3`

`=>3r=3+9=12`

`=>r=12/3`

`=>r=4`

Thus, total number of students `(s)=rxxn`

`=4xx9=36`

Thus, total number of students in the class = 36 Answer

Question: 5. In a `triangle\ ABC`, `/_C=3 /_B=2(/_A+/_B)`. Find the three angles.

Soltuion:

Given,

`/_C=3 /_B=2(/_A+/_B)`

`=>3/_B=2/_A+2/_B`

`=>3/_B-2/_B=2/_A`

`=>/_B=2/_A` ---------- (i)

Now, we know that,

In a triangle, sum of all the three agnels `=180^o`

Thus, in `triangleABC`

`/_A+/_B+/_C=180^o`

`=>/_A+/_B+3/_B=180^o`

[∵ as given in question `/_C=3/_B`]

`=>/_A+4/_B=180^o`

After substituting the value `/_B=2/_A` from equation (i), we get

`=>/_A+4(2/_A)=180^o`

`=>/_A+8/_A=180^0`

`=>9/_A=180^o`

`=>/_A=(180^o)/9`

`=>/_A=20^o`

Now, by substituting the value `/_A=20^o` in equation (i) [`/_B=2/_A`], we get

`/_B=2xx20^o`

`=>/_B=40^0`

Now, since, as given in question `/_C=3/_B` by substituting the value `/_B=40^o`, we get

`/_C=3xx40^o`

`=>/_C=120^o`

Thus, `/_A=20^o, /_B=40^o` And `/_C=120^o` Answer

Question: 6. Draw the graphs of the equations `5x-y=5` and `3x-y=3`. Determine the co–ordinates of the vertices of the triangle formed by these lines and the `y`–axis.

Solution:

Given, pair of linear equations

`5x-y=5` ; and `3x-y=3`

Now, from `5x-y=5`

`=>5x=5+y`

`=>x=(5+y)/5`

Thus, table of solution for this equation:

`x`	0	1	2
`y`	–5	0	5

And from equation, `3x-y=3`

`=>3x=3+y`

`=>x=(3+y)/3`

Table of solution for this equation:

`x`	0	1	2
`y`	–3	0	3

Thus, coordinates of triangle formed by `y`–axis are (1, 0), (0, –5) and (0, –3) Answer

Question: 7. Solve the following pair of linear equation:

(i) `px+qy=p-q`

`qx-py=p+q`

Solution:

Given, `px+qy=p-q` -------- (i)

And, `qx-py=p+q` -------- (ii)

After multiplying equation (i) by`p` we get

`p(px+qy=p-q)`

`=>p^2x+pqy=p^2-pq` ----------- (iii)

After multiplying equation (ii) by `q` we get

`q(qx-py=p+q)`

`=>q^2x-qpy=pq+q^2` --------------- (iv)

Now, by adding equation (iii) and (iv), we get

`(p^2x+pqy)+(q^2x-pqy)` `=(p^2-pq)+(pq+q^2)`

`=>p^2x+\cancel(pqy)+q^2x-\cancel(pqy)` `=p^2-\cancel(pq)+\cancel(pq)+q^2`

`=>p^2x+q^x=p^2+q^2`

`=>x(p^2+q^2)=p^2+q^2`

`=>x=(p^2+q^2)/(p^2+q^2)`

`=>x=1`

Now, by substituting, the value `x=1` in equation (i), we get

`pxx1+qy=p-q`

`=>p+qy=p-q`

`=>qy=p-q-p`

`=>qy = -q`

`=>y=(-q)/q`

`=>y=-1`

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

(ii) `ax+by=c`

`bx+ay=1+c`

Given, `ax+by=c` ----------- (i)

`bx+ay=1+c` ----------- (ii)

After multiplying equation (i) by `a`, we get

`a(ax+by=c)`

`=>a^2x+aby=ac` ---------- (iii)

After multiplying equation (ii) by `b`, we get

`b(bx+ay=1+c)`

`=>b^2x+aby=b+bc` ------------ (iv)

Now, equation (iii) `– equation (iv), we get

`a^2x+aby-(b^2x+aby)=ac-(b+bc)`

`=>a^2x+\cancel(aby)-b^2x-\cancel(aby)=ac-b-bc`

`=>x(a^2-b^2)=c(a-b)-b`

`=>x=(c(a-b)-b)/(a^2-b^2)` ---------- (v)

Now, by substituting the value of `x` in equation (i) [`ax+by=c`], we get

`a{(c(a-b)-b)/(a^2-b^2)}+by=c`

`=>(ac(a-b)-ab)/(a^2-b^2)+by=c`

`=>by=c-(ac(a-b)-ab)/(a^2-b^2)`

`=>by=(c(a^2-b^2)-(ac(a-b)-ab))/(a^2-b^2)`

`=>by=(a^2c-b^2c-(a^2c-acb-ab))/(a^2-b^2)`

`=>by=(\cancel(a^2c)-b^2c-\cancel(a^2c)+abc+ab)/(a^2-b^2)`

`=>by=(abc-b^2c+ab)/(a^2-b^2)`

`=>by=(bc(a-b)+ab)/(a^2-b^2)`

`=>by=(b(c(a-b)+a))/(a^2-b^2)`

`=>y=(b(c(a-b)+a))/(b(a^2-b^2))`

`=>y=(c(a-b)+a)/(a^2-b^2)`

Thus, `x=(c(a-b)-b)/(a^2-b^2)` and

` y=(c(a-b)+a)/(a^2-b^2)` Answer

(iii) `x/a-y/b=0`

`ax+by=a^2+b^2`

Solution:

Given, `x/a-y/b=0` ----------- (i)

`ax+by=a^2+b^2` --------- (ii)

From equation (i)

`x/a-y/b=0`

`=>(bx-ay)/(ab)=0`

After cross multiplication

`=>bx-ay=0` ------- (iii)

After multiplying by `b`, we get

`b(bx-ay=0)`

`=>b^2x-aby=0` ---------- (iv)

Now, after multiplying equation (ii) by `a`, we get

`a(ax+by=a^2+b^2)`

`=>a^2x+aby=a^3+ab^2` -------- (v)

Now, adding equation (iv) and (v)

`b^2x-\cancel(aby)+a^2x+\cancel(aby)=a^3+ab^2`

`=>b^2x+a^2x=a^3+ab^2`

By taking `x` from LHS and `a` from RHS as common

`=>x(b^2+a^2)=a(a^2+b^2)`

`=>x=a(a^2+b^2)/(a^2+b^2)`

`=>x=a`

Now, after substituting the value of `x=a` in equation (iii), we get

`=>ba-ay=0`

`=>ay = ba`

`=>y=(ba)/a`

`=>y=b`

Thus, `x=a` and `y=b` Answer

(iv) `(a-b)x+(a+b)y=a^2-2ab-b^2`

`(a+b)(x+y)=a^2+b^2`

Solution:

Given,

`(a-b)x+(a+b)y=a^2-2ab-b^2` ---------- (i)

`(a+b)(x+y)=a^2+b^2` ----------- (ii)

`=>(a+b)x+(a+b)y=a^2+b^2` ---------- (iii)

After subtracting equation (iii) from equation (i), we get

`=>-2bx=-2b(a+b)`

`=>x=(-2b(a+b))/(-2b)`

`=>x = a+b`

Now, by substituting `x=a+b` in equation (i), we get

`=>(a+b)y=-2ab`

`=>y=(-2ab)/(a+b)`

Thus, `x = (a+b)` and `y=(-2ab)/(a+b)` Answer

(v) `152x-378y=-74`

`-378x+152y=-604`

Solution:

Given, `152x-378y=-74` ---------- (i)

`-378x+152y=-604` ---------- (ii)

After multiplying equation (i) by 378, we get

`378(152x-378y=-74)`

`=>57456x-142884y=-27972` -------- (iii)

After multiplying equation (ii) by 152, we get

`152(-378x+152y=-604)`

`=>-57456x+23104y=-91808` --------- (iv)

Now, by adding equation (iii) and (iv), we get

`=>-119780y=-119780`

`=>y=(-119780)/(-119780)`

`=>y=1`

Now, by substituting, value of `y=1` in equation (i), we get

`152x-378xx1=-74`

`=>152x-378=74`

`=>152x=-74+378`

`=>152x=304`

`=>x = 304/152`

`=>x=2`

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

Question: 8. ABCD is acyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:

Given, For cyclic quadrilateral

`/_A = 4y+20`

`/_B=3y-5`

`/_C=-4x`

`/_D=-7x+5`

Now, we know that sum of the opposite angels of a quadrilateral is equal to `180^o`

Here, `/_A` and `/_C` are opposite angles.

Thus, `/_A+/_C=180^o`

After substituting the value of `/_A` and `/_C` , we get

`4y+20+(-4x)=180`

`=>4y-4x=180-20`

By taking 4 as common from LHS

`=>4(y-x)=160`

`=>y-x=160/4`

`=>y-x=40` ------- (i)

Now, `/_B` and `/_D` are opposite angles,

Thus, `/_B+/_D=180`

[∵ sum of opposite angles of a quadrilateral = 180]

After substituting the values of `/_B` and `/_D`, we get

`3y-5+(-7x+5)=180`

`=>3y-5-7x+5=180`

`=>3y-7x=180` --------- (ii)

Now, after multiplying equation (i) by 3, we get

`3(y-x=40)`

`=>3y-3x=120` ------- (iii)

Now, after subtracting equation (iii) from equation (ii), we get

`(3y-7x)-(3y-3x)=180-120`

`=>\cancel(3y)-7x-\cancel(3y)+3x=60`

`=>-4x=60`

`=>x=60/(-4)=-15`

Now, by substituting the value of `x=-15` in equation (i) [`y-x=40`], we get

`y-(-15)=40`

`=>y+15=40`

`=>y=40-15=25`

Now, as given,

`/_A=4y+20`

Thus, by substituting value of `y=25`, we get

`/_A = 4xx25+20=100+20`

`=>/_A=120^o`

And as given, `/_B=3y-5`

Thus, by substituting the value of `y=25`, we get

`/_B=3xx25-5=75-5`

`=>/_B=70^o`

And, as given, `/_C=-4x`

Thus, by substituting the value of `x=-15`, we get

`/_C=-4xx(-15)=60^o`

And, as given, `/_D=-7x+5`

Thus, by substituting the value of `x=-15`, we get

`/_D=-7xx(-15)+5 =105+5`

`=>/_D=110^o`

Thus, `/_A=120^o, /_B=70^o`, `/_C=60^o` and `/_D=110^o` Answer

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