The two lines intersect at the point (1.0). So, r= 1, y=0 is the required solution
e pair of linear equations, ie, the number of pants she purchased is 1 and she did
uy any skirt.
y the answer by checking whether it satisfies the conditions of the given
em
EXERCISE 3.2
Form the pair of linear equations in the following problems, and find their solution
graphically.
(1) 10 students of Class X took part in a Mathematics quiz. If the number of girls is
more than the number of boys, find the number of boys and girls who took part
the quiz
(ii) 5 pencils and 7 pens together costi 50, whereas 7 pencils and 5 pens togeth
cost 46. Find the cost of one pencil and that of one pen.
Answers
Answer:
Correct equations :-
\sf{\:\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=\dfrac{1}{2}}
16x+24z
4
+
21x−14z
12
=
2
1
\sf{\:\dfrac{14}{4{\bf{x}}+6z}+\dfrac{4}{3x-2z}=2}
4x+6z
14
+
3x−2z
4
=2
Solution :-
Given first equation:
\sf\implies\dfrac{4}{16x+24z}+\dfrac{12}{21x-14z}=1/2⟹
16x+24z
4
+
21x−14z
12
=1/2
\sf\implies\dfrac{4}{4(x+6z)}+\dfrac{12}{7(3x-2z)}=1/2⟹
4(x+6z)
4
+
7(3x−2z)
12
=1/2
\sf\implies\dfrac{1}{4x+6z}+\dfrac{12}{7(3x-2z)}=1/2⟹
4x+6z
1
+
7(3x−2z)
12
=1/2
Assume that :-
4x + 6z = u
3x - 2z = v
\sf\implies\dfrac{1}{u}+\dfrac{12}{7v}=1/2⟹
u
1
+
7v
12
=1/2
\sf\implies\dfrac{7v+12u}{7vu}=1/2⟹
7vu
7v+12u
=1/2
\sf\implies 14v + 24 u = 7vu---(1.)⟹14v+24u=7vu−−−(1.)
Given second equation:
\sf\implies\dfrac{14}{4x+6z}+\dfrac{4}{3x-2z}=2⟹
4x+6z
14
+
3x−2z
4
=2
\sf\implies\dfrac{14}{u}+\dfrac{4}{v}=2⟹
u
14
+
v
4
=2
\sf\implies\dfrac{14v+4u}{vu}=2⟹
vu
14v+4u
=2
\sf\implies{14v+4u}=2vu---(2.)⟹14v+4u=2vu−−−(2.)
Eq(1) - Eq(2)
\implies⟹ 14v+24u - (14 v + 4 u) =7vu-2vu
\implies⟹ 24u - 4u = 5vu
\implies⟹ 20 u = 5vu
\implies⟹ 4 = v
Put v in eq(2)
\implies⟹ 14(4) + 4u = 2(4)u
\implies⟹ 56 + 4u = 8u
\implies⟹ 56 = 8u - 4u
\implies⟹ 56 = 4u
\implies⟹ 56/4 = u
\implies⟹ 14 = u
Now we get :-
u = 14 = 4x + 6z
v = 4 = 3x - 2z ---(3)
Multiply equation (3) with 3
\implies⟹ 3(4 = 3x -2z)
\implies⟹ 12 = 9x - 6z ---(4)
Add u and eq(4)
\implies⟹ 14+12 = 4x + 6z + 9x - 6z
\implies⟹ 26 = 13x
\implies⟹ 26/13 = x
\implies⟹ 2 = x
So the value of x is 2.
Put this value of x in u
\implies⟹ 14 = 4(2) + 6z
\implies⟹ 14 = 8 + 6z
\implies⟹ 14 - 8 = 6z
\implies⟹ 6 = 6z
\implies⟹ 1 = z
So the value of z is 1.