Exercise 3.4
Solve each of the following systems of equation by the method of cross-multiplication:
1. x + 2y + 1 = 0 and 2x – 3y – 12 = 0
Solution:
Given equations are:
x+2y+1 =0 …………………………………….(i)
2x-3y-12=0………………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 1 , b1= 2 , c1= 1
a2= 2 , b2= -3 , c2= -12
By cross multiplication method,
\(\frac{x}{-24+3}=\frac{-y}{-12-2}=\frac{1}{-3-4}\)
\(\frac{x}{-21}=\frac{-y}{-14}=\frac{1}{-7}\)
Now,
\(\frac{x}{-21}=\frac{1}{-7}\)
=x= 3
And,
\(\frac{-y}{-14}=\frac{1}{-7}\)
=y=-2
∴, the solution of the given system of equation is x=3 and y=-2 respectively.
2. 3x + 2y + 25 = 0, 2x + y + 10 = 0
Solution:
Given equations are:
3x+2y+25 =0 …………………………………….(i)
2x+y+10=0………………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 3 , b1= 2 , c1= 25
a2= 2 , b2= 1 , c2= 10
By cross multiplication method,
\(\frac{x}{20-25}=\frac{-y}{30-50}=\frac{1}{3-4}\)
\(\frac{x}{-5}=\frac{-y}{-20}=\frac{1}{-1}\)
Now,
\(\frac{x}{-5}=\frac{1}{-1}\)
=x= 5
And,
\(\frac{-y}{-20}=\frac{1}{-1}\)
=y=-20
∴, the solution of the given system of equation is x=5 and y=-20 respectively.
3. 2x + y = 35, 3x + 4y = 65
Solution:
Given equations are
2x+y= 35 ……………………………….(i)
3x+4y=65…………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 2 , b1= 1 , c1= -35
a2= 3 , b2= 4 , c2= -65
By cross multiplication method,
\(\frac{x}{-65+140}=\frac{-y}{-130+105}=\frac{1}{8-3}\)
\(\frac{x}{75}=\frac{-y}{-25}=\frac{1}{5}\)
Now,
\(\frac{x}{75}=\frac{1}{5}\)
=x= 15
And,
\(\frac{-y}{-25}=\frac{1}{5}\)
=y=5
∴, the solution of the given system of equation is x=15 and y=5 respectively.
4. 2x – y – 6 = 0, x – y – 2 = 0
Solution:
Given equations are:
2x-y= 6 ……………………………….(i)
x-y=2…………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 2 , b1= -1 , c1= -6
a2= 1 , b2= -1 , c2= -2
By cross multiplication method,
\(\frac{x}{2-6}=\frac{-y}{-4+6}=\frac{1}{-2+1}\)
\(\frac{x}{-4}=\frac{-y}{2}=\frac{1}{-1}\)
Now,
\(\frac{x}{-4}=\frac{1}{-1}\)
=x= 4
And,
\(\frac{-y}{2}=\frac{1}{-1}\)
=y=2
∴, the solution of the given system of equation is x=4 and y=2 respectively.
5. \(\frac{x+y}{xy}=2\), \(\frac{x-y}{xy}=6\)
Solution:
Given equations are:
\(\frac{x+y}{xy}=2\)
= \(\frac{1}{x}+\frac{1}{y}=2\) ……………………………….. (i)
\(\frac{x-y}{xy}=6\)
= \(\frac{1}{x}-\frac{1}{y}=6\) ………………………………… (ii)
Taking \(\frac{1}{x}\) =u
Taking \(\frac{1}{y}\) =v
= u+v=2 …………………………… (iii)
= u-v= 6 ……………………………. (iv)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 1 , b1= 1 , c1= -2
a2= 1 , b2= -1 , c2= -6
By cross multiplication method,
\(\frac{u}{6-2}=\frac{-v}{6+2}=\frac{1}{-1-1}\)
\(\frac{u}{4}=\frac{-v}{8}=\frac{1}{-2}\)
Now,
\(\frac{u}{4}=\frac{1}{-2}\)
=u= -2
And,
\(\frac{-v}{-8}=\frac{1}{-2}\)
=v=4
\(\frac{1}{u}\) =x=\(\frac{-1}{2}\)
\(\frac{1}{v}\) =y=\(\frac{1}{4}\)
∴, the solution of the given system of equation is x=\(\frac{-1}{2}\) and y=\(\frac{1}{4}\) respectively.
6. ax+by=a-b, bx-ay=a+b
Solution:
Given equations are:
ax+by=a-b……………………………….(i)
bx-ay=a+b…………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= a , b1= b , c1= -(a-b)
a2= b , b2= -a , c2= -(a+b)
By cross multiplication method,
\(\frac{x}{-ab-b^{2}+ab-a^{2}}=\frac{-y}{-a^{2}-ab-b^{2}+ab}=\frac{1}{-a^{2}-b^{2}}\)
\(\frac{x}{-b^{2}-a^{2}}=\frac{-y}{-a^{2}-b^{2}}=\frac{1}{-a^{2}-b^{2}}\)
Now,
\(\frac{x}{-ab-b^{2}+ab-a^{2}}=\frac{1}{-a^{2}-b^{2}}\)
=x= 1
And,
\(\frac{-y}{-a^{2}-ab-b^{2}+ab}=\frac{1}{-a^{2}-b^{2}}\)
=y=-1
∴, the solution of the given system of equation is x=1 and y=-1 respectively.
7. x+ay-b=0, ax-by-c=0
Solution:
Given equations are:
x+ay-b=0 ……………………………….. (i)
ax-by-c=0………………………………. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 1 , b1= a , c1= -b
a2= a , b2= -b , c2= -c
By cross multiplication method,
\(\frac{x}{-ac-b^{2}}=\frac{-y}{-c+ab}=\frac{1}{-a^{2}-b}\)
Now,
\(\frac{x}{-ac-b^{2}}=\frac{1}{-a^{2}-b}\)
=x= \(\frac{b^{2}+ac}{a^{2}+b}\)
And,
\(\frac{-y}{-c+ab}=\frac{1}{-a^{2}-b}\)
=y=\(\frac{-c+ab}{a^{2}+b}\)
∴, the solution of the given system of equation is x=\(\frac{b^{2}+ac}{a^{2}+b}\) and y=\(\frac{-c+ab}{a^{2}+b}\) respectively.
8.ax+by=a2 and bx+ay=b2
Solution:
Given equations are:
ax+by=a2……………………………….(i)
bx+ay=b2…………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= a , b1= b , c1= a2
a2= b , b2= a , c2= b2
By cross multiplication method,
\(\frac{x}{-b^{2}+a^{2}}=\frac{-y}{-ab^{2}-a^{2}b}=\frac{1}{a^{2}-b^{2}}\)
Now,
\(\frac{x}{-b^{2}+a^{2}}=\frac{1}{a^{2}-b^{2}}\)
=x= \(\frac{a^{2}+ab+b^{2}}{a+b}\)
And,
\(\frac{-y}{-ab^{2}-a^{2}b}=\frac{1}{a^{2}-b^{2}}\)
=y=-\(\frac{-ab(a-b)}{(a-b)(a+b)}\)
∴, the solution of the given system of equation is x=\(\frac{a^{2}+ab+b^{2}}{a+b}\) and y=\(\frac{-ab(a-b)}{(a-b)(a+b)}\) respectively.
9.
\(\frac{5}{x+y}-\frac{2}{x-y}=-1\)
\(\frac{15}{x+y}+\frac{7}{x-y}=-10\)
Solution:
Given equations are:
\(\frac{5}{x+y}-\frac{2}{x-y}=-1\)
\(\frac{15}{x+y}+\frac{7}{x-y}=-10\)
Let \(\frac{1}{x+y}=\) =u
Let \(\frac{1}{x-y}=\) =v
The given system of equations are :
5u-2v=-1
15u+7v =10
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 5, b1= -2 , c1= 1
a2= 15 , b2= 7 , c2= -10
By cross multiplication method,
\(\frac{u}{20-7}=\frac{-v}{-50-15}=\frac{1}{35+30}\)
\(\frac{u}{13}=\frac{-v}{-65}=\frac{1}{65}\)
Now,
\(\frac{u}{13}=\frac{1}{-65}\)
=u= \(\frac{1}{5}\)
\(\frac{1}{u}=\) =x+y
=x+y=5 ………………………..(i)
And,
\(\frac{-v}{-65}=\frac{1}{-65}\)
=v=1
\(\frac{1}{v}=\) =x-y
=x-y =1 ………………………… (ii)
Adding equation (i) and (ii)
2x= 6
=x=3
Substituting the value of x in equation (i)
3+y=5
=y=2
The solution of the given system of equation is 3 and 2 respectively.
10.
\(\frac{2}{x}+\frac{3}{y}=13\)
\(\frac{5}{x}-\frac{4}{y}=-2\)
Solution:
Given equations are:
\(\frac{2}{x}+\frac{3}{y}=13\)
\(\frac{5}{x}-\frac{4}{y}=-2\)
Let \(\frac{1}{x}\) =u
Let \(\frac{1}{y}\) =v
The given system of equations becomes:
2u+3v=13 ……………………………… (i)
5u-4v=-2…………………………………. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 2, b1= 3 , c1= -13
a2= 5 , b2= -4 , c2= 2
By cross multiplication method,
\(\frac{u}{6-52}=\frac{-v}{4+65}=\frac{1}{-8-15}\)
\(\frac{u}{-46}=\frac{-v}{69}=\frac{1}{-23}\)
Now,
\(\frac{u}{-46}=\frac{1}{-23}\)
=u= 2
\(\frac{1}{u}=\) =\(\frac{1}{x}\)
=x=\(\frac{1}{2}\)
And,
\(\frac{-v}{69}=\frac{1}{-23}\)
=v=3
\(\frac{1}{v}\)= \(\frac{1}{y}\)
=y =\(\frac{1}{3}\)
∴, the solutions of the given system of equations are x=\(\frac{1}{2}\) and y=\(\frac{1}{3}\) respectively.
11.
\(\frac{57}{x+y}+\frac{6}{x-y}=5\)
\(\frac{38}{x+y}+\frac{21}{x-y}=9\)
Solution:
Given equations are:
\(\frac{57}{x+y}+\frac{6}{x-y}=5\)
\(\frac{38}{x+y}+\frac{21}{x-y}=9\)
Let \(\frac{1}{x+y}=\) =u
Let \(\frac{1}{x-y}=\) =v
The given system of equations are :
57u+6v=5
38u+21v =9
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 57, b1= 6 , c1= -5
a2= 38 , b2=21 , c2= -9
By cross multiplication method,
\(\frac{u}{-54+105}=\frac{-v}{-513+190}=\frac{1}{1193-228}\)
\(\frac{u}{51}=\frac{-v}{-323}=\frac{1}{969}\)
Now,
\(\frac{u}{51}=\frac{1}{969}\)
=u= \(\frac{1}{19}\)
\(\frac{1}{u}\) =x+y
=x+y=19 ………………………..(i)
And,
\(\frac{-v}{-323}=\frac{1}{969}\)
=v=\(\frac{1}{3}\)
\(\frac{1}{v}=\) =x-y
=x-y =3 ………………………… (ii)
Adding equation (i) and (ii)
2x= 22
=x=11
Substituting the value of x in equation (i)
11+y=19
=y=8
∴, the solution of the given system of equation is x=11 and y=8 respectively.
12.
\(\frac{x}{a}-\frac{y}{b}=2\)
ax-by=a2-b2
Solution:
Given equations are:
\(\frac{x}{a}-\frac{y}{b}=2\)
ax-by=a2-b2
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1=\(\frac{1}{a}\) , Let b1=\(\frac{1}{b}\), Let c1=-2
a2=a , b2=-b,  c2=b2-a2
By cross multiplication method
= \(\frac{x}{\frac{b^{2}-a^{2}}{b}-2b}=\frac{-y}{\frac{b^{2}-a^{2}}{b}+2b}=\frac{1}{\frac{-b}{a}-\frac{a}{b}}\)
= \(\frac{x}{\frac{b^{2}-a^{2}-2b^{2}}{b}}=\frac{-y}{\frac{b^{2}-a^{2}+2b^{2}}{b}}=\frac{1}{\frac{-b^{2}-a^{2}}{ab}}\)
Now, \(\frac{x}{\frac{b^{2}-a^{2}-2b^{2}}{b}}=\frac{1}{\frac{-b^{2}-a^{2}}{ab}}\)
x= a
and, \(\frac{-y}{\frac{b^{2}-a^{2}+2b^{2}}{b}}=\frac{1}{\frac{-b^{2}-a^{2}}{ab}}\)
=y=b
∴, the solution of the given system of equation are x=a and y=b respectively.
13.
\(\frac{x}{a}+\frac{y}{b}=a+b\)
\(\frac{x}{a^{2}}+\frac{y}{b^{2}}=2\)
Solution:
Given equations are:
\(\frac{x}{a}+\frac{y}{b}=a+b\)
\(\frac{x}{a^{2}}+\frac{y}{b^{2}}=2\)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1=\(\frac{1}{a}\) , Let b1=\(\frac{1}{b}\), Let c1=-(a+b)
a2=\(\frac{1}{a^{2}}\), b2=\(\frac{1}{b^{2}}\),  c2=-2
By cross multiplication method
=\(\frac{x}{\frac{-2}{b}+\frac{a}{b^{2}}+\frac{1}{b}}=\frac{-y}{\frac{-2}{a}+\frac{1}{a}+\frac{b}{a^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)
= \(\frac{x}{\frac{a-b}{b^{2}}}=\frac{-y}{\frac{-a-b}{a^{2}}+\frac{1}{a}+\frac{b}{a^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)
Now, \(\frac{x}{\frac{a-b}{b^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)
=x=a2
\(\frac{-y}{\frac{-a-b}{a^{2}}+\frac{1}{a}+\frac{b}{a^{2}}}=\frac{1}{\frac{-1}{ab^{2}}-\frac{-1}{a^{2}b}}\)
=y=b2
∴, the solution of the given system of equation are x=a2 and y=b2 respectively.
14.
\(\frac{x}{a}=\frac{y}{b}\)
ax+by=a2+b2
Solution:
Given equations are:
\(\frac{x}{a}=\frac{y}{b}\)
ax+by=a2+b2
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1=\(\frac{1}{a}\) , Let b1=\(\frac{1}{b}\), Â c1=0
Hence, a1=a , b2=b, Let c1=-(a2+b2)
By cross multiplication method
\(\frac{x}{\frac{a^{2}+b^{2}}{b}}= \frac{y}{\frac{a^{2}+b^{2}}{a}}=\frac{1}{\frac{a}{b}+\frac{b}{a}} \)
Now, \(\frac{x}{\frac{a^{2}+b^{2}}{b}}=\frac{1}{\frac{a}{b}+\frac{b}{a}} \)
=x=a
And \( \frac{y}{\frac{a^{2}+b^{2}}{a}}=\frac{1}{\frac{a}{b}+\frac{b}{a}} \)
=y=b
∴, the solution of the given system of equations are x=a and y=b respectively.
15.
2ax+3by=a+2b
3ax+2by=2a+b
Solution:
Given equations are:
2ax+3by=a+2b …………………………… (i)
3ax+2by=2a+b…………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 2a, b1= 3b , c1= -(a+2b)
a2= 3a , b2=2b , c2= -(2a+b)
By cross multiplication method
\(\frac{x}{-4ab+b^{2}}=\frac{-y}{-a^{2}+4ab}=\frac{1}{-5ab}\)
Now,
\(\frac{x}{-4ab+b^{2}}=\frac{1}{-5ab}\)
=x= \(\frac{4a-b}{5a}\)
And, \(\frac{-y}{-a^{2}+4ab}=\frac{1}{-5ab}\)
=y=\(\frac{4b-a}{5b}\)
∴, the solutions of the system of equations are x= \(\frac{4a-b}{5a}\) and y=\(\frac{4b-a}{5b}\).
16.Â
5ax+6by=28
3ax+4by=18
Solution:
Given equations are:
5ax+6by=28 …………………………. (i)
3ax+4by=18……………………………. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 5a, b1= 6b , c1= -(28)
a2= 3a , b2=4b , c2= -(18)
By cross multiplication method
\(\frac{x}{4b}=\frac{-y}{-6a}=\frac{1}{2ab}\)
Now,
\(\frac{x}{4b}=\frac{1}{2ab}\)
=x= \(\frac{2}{a}\)
And, \(\frac{-y}{-6a}=\frac{1}{2ab}\)
=y=\(\frac{3}{b}\)
∴, the solution of the given system of equation is x=\(\frac{2}{a}\) and y=\(\frac{3}{b}\).
17.
(a+2b)x+(2a-b)y=2
(a-2b)x+(2a+b)y=3
Solution:
Given equations are :
(a+2b)x+(2a-b)y=2 ………………………. (i)
(a-2b)x+(2a+b)y=3………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= a+2b, b1= 2a-b , c1= -(2)
a2= a-2b , b2=2a+b , c2= -(3)
By cross multiplication method:
\(\frac{x}{-2a+5b}=\frac{y}{a+10b}=\frac{1}{10ab}\)
Now, \(\frac{x}{-2a+5b}=\frac{1}{10ab}\)
=x= Â \(\frac{5b-2a}{10ab}\)
And \(\frac{y}{a+10b}=\frac{1}{10ab}\)
=y=\(\frac{a+10b}{10ab}\)
∴, the solution of the system of equations are x= \(\frac{5b-2a}{10ab}\)
And y=\(\frac{a+10b}{10ab}\) respectively.
18.
\(x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})\)
x+y=2a2
Solution:
Given equations are:
\(x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})\)
x+y=2a2
From equation (i)
\(x(\frac{a^{2}+b^{2}-2ab+ab}{a-b})-y(\frac{a^{2}+b^{2}+2ab-ab}{a+b})\)
= \(x(\frac{a^{2}+b^{2}-ab}{a-b})-y(\frac{a^{2}+b^{2}+ab}{a+b})\) ……………….. (iii)
From equation (ii)
x+y-2a2=0
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= \((\frac{a^{2}+b^{2}-ab}{a-b})\) , b1= –\((\frac{a^{2}+b^{2}+ab}{a-b})\) , c1=0
a2=1, b2=1 , c2=-2a2
By cross multiplication method:
\(\frac{x}{2a^{2}(\frac{a^{2}+b^{2}+ab}{a+b})}=\frac{-y}{(-2a^{2})(\frac{a^{2}+b^{2}-ab}{a-b})}=\frac{1}{\frac{2a^{3}}{(a-b)(a+b)}}\)
Now, \(\frac{x}{2a^{2}(\frac{a^{2}+b^{2}+ab}{a+b})}=\frac{1}{\frac{2a^{3}}{(a-b)(a+b)}}\)
=x=\(\frac{a^{3}-b^{3}}{a}\)
And \(\frac{-y}{(-2a^{2})(\frac{a^{2}+b^{2}-ab}{a-b})}=\frac{1}{\frac{2a^{3}}{(a-b)(a+b)}}\)
=y=\(\frac{a^{3}+b^{3}}{a}\)
∴, the solutions of the given system of equations are x= \(\frac{a^{3}-b^{3}}{a}\) and y=\(\frac{a^{3}+b^{3}}{a}\) respectively.
19.
bx+cy=a+b
\(-ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}+\frac{1}{b+a})=\frac{2a}{a+b}\)
Solution:
Given equations are:
bx+cy=a+b ……………………………………. (i)
\(-ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}+\frac{1}{b+a})=\frac{2a}{a+b}\)…..(ii)
From equation (i)
bx+cy-(a+b) =0
From equation (ii)
\(-ax(\frac{1}{a-b}-\frac{1}{a+b})+cy(\frac{1}{b-a}+\frac{1}{b+a})-\frac{2a}{a+b}=0\)
= \(x(\frac{2ab}{(a-b)(a+b)})+y(\frac{2ac}{(b-a)(b+a)})-\frac{2a}{a+b}=0\)
= \(\frac{1}{a+b}(\frac{2abx}{a-b}-\frac{2acy}{a-b}-2a)=0\)
= \(\frac{2abx}{a-b}-\frac{2acy}{a-b}-2a=0\)
= 2abx-2acy-2a(a-b) =0 …………………………. (iv)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= b, b1= c , c1=-(a+b)
a2=2ab, b2=-2ac , c2=-2a(a-b)
By cross multiplication
= \(\frac{x}{-4a^{2}c}=\frac{-y}{4ab^{2}}=\frac{-1}{4abc}\)
Now, \(\frac{x}{-4a^{2}c}=\frac{-1}{4abc}\)
=x=\(\frac{a}{b}\)
And,
= \(\frac{-y}{4ab^{2}}=\frac{-1}{4abc}\)
=y=\(\frac{b}{c}\)
∴, the solution of the system of equations are x=\(\frac{a}{b}\) and y=\(\frac{b}{c}\)
20.
(a-b)x+(a+b)y=2a2-2b2
(a+b)(x+y) =4ab
Soln.
Given equations are :
(a-b)x+(a+b)y=2a2-2b2 ………………………….. (i)
(a+b)(x+y) =4ab …………………………. (ii)
From equation (i)
(a-b)x+(a+b)y-2a2-2b2 =0
= (a-b)x+(a+b)y-2(a2-b2) =0
From equation (ii)
(a-b)x+(a-b)y-4ab=0
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= a-b , b1 = a+b , c1=-2(a2+b2)
a2= a+b , b2 = a+b , c2=-4ab
By cross multiplication method
\(\frac{x}{2(a+b)(a^{2}-b^{2}+2ab)}=\frac{-y}{2(a-b)(a^{2}+b^{2})}=\frac{1}{-2b(a+b)}\)
Now,
\(\frac{x}{2(a+b)(a^{2}-b^{2}+2ab)}=\frac{1}{-2b(a+b)}\)
=x= \(\frac{2ab-a^{2}+b^{2}}{b}\)
And, \(\frac{-y}{2(a-b)(a^{2}+b^{2})}=\frac{1}{-2b(a+b)}\)
=y=\(\frac{(a-b)(a^{2}+b^{2})}{b(a+b)}\)
∴, the solution of the system of equations are x= \(\frac{2ab-a^{2}+b^{2}}{b}\) and y=\(\frac{(a-b)(a^{2}+b^{2})}{b(a+b)}\) respectively.
21.
a2x+b2y=c2
b2x+a2y=d2
Solution:
The given system of equations are :
a2x+b2y=c2 ………………………………….. (i)
b2x+a2y=d2……………………………………… (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= a2 , b1 = b2 , c1=-c2
a2= b2 , b2 = a2 , c2=-d2
By cross multiplication method
= \(\frac{x}{-b^{2}d^{2}+a^{2}c^{2}}=\frac{-y}{-a^{2}d^{2}+b^{2}c^{2}}=\frac{1}{a^{4}-b^{4}}\)
Now,
\(\frac{x}{-b^{2}d^{2}+a^{2}c^{2}}=\frac{1}{a^{4}-b^{4}}\)
=x= \(\frac{a^{2}c^{2}-b^{2}d^{2}}{a^{4}-b^{4}}\)
And, = \(\frac{x}{-b^{2}d^{2}+a^{2}c^{2}}=\frac{-y}{-a^{2}d^{2}+b^{2}c^{2}}=\frac{1}{a^{4}-b^{4}}\)
=y= \(\frac{a^{2}d^{2}-b^{2}c^{2}}{a^{4}-b^{4}}\)
∴, the solution of the given system of equations are x=\(\frac{a^{2}c^{2}-b^{2}d^{2}}{a^{4}-b^{4}}\) and y=\(\frac{a^{2}d^{2}-b^{2}c^{2}}{a^{4}-b^{4}}\) respectively.
22.Â
\(ax+by=\frac{a+b}{2}\)
3x+5y=4
Solution:
Given equations are:
\(ax+by=\frac{a+b}{2}\)3x+5y=4
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 2a , b1 = 2b , c1= -(a+b)
a2= 3 , b2 = 5 , c2= -4
By cross multiplication method
23.
2(ax-by)+a+4b=0
2(bx+ay)+b-4a=0
Solution:
Given equations are:
2(ax-by)+a+4b=0 …………………….. (i)
2(bx+ay)+b-4a=0……………………. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 2a , b1 = -2b , c1=a+4b
a2= 2b , b2 = 2a , c2=b-4a
By cross multiplication method
=\(\frac{x}{-2b^{2}+8ab-2a^{2}-8ab}=\frac{-y}{2ab-8a^{2}-2ab-8b^{2}}=\frac{1}{4a^{2}+4b^{2}}\)
= \(\frac{x}{-2b^{2}-2a^{2}}=\frac{-y}{-8a^{2}-8b^{2}}=\frac{1}{4a^{2}+4b^{2}}\)
Now, \(\frac{x}{-2b^{2}-2a^{2}}=\frac{1}{4a^{2}+4b^{2}}\)
=x= \(\frac{-1}{2}\)
And, \(\frac{-y}{-8a^{2}-8b^{2}}=\frac{1}{4a^{2}+4b^{2}}\)
=y=2
∴, the solution of the given pair of equations are x=\(\frac{-1}{2}\) and y=2 respectively.
24.
6(ax+by)=3a+2b
6(bx-ay) = 3b-2a
Solution:
Given equations are
6(ax+by)=3a+2b ………………………… (i)
6(bx-ay) = 3b-2a …………………………. (ii)
From equation (i)
6ax+6by-(3a+2b)=0 ……………………… (iii)
From equation (ii)
6bx-6ay-(3b-2a) =0 …………………………… (iv)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= 6a , b1 = 6b , c1=-(3a+2b)
a2= 6b , b2 = -6a , c2=-(3b-2a)
By cross multiplication method
\(\frac{x}{-18(a^{2}+b^{2})}=\frac{-y}{12(a^{2}+b^{2})}=\frac{-1}{36(a^{2}+b^{2})}\)
Now, \(\frac{x}{-18(a^{2}+b^{2})}=\frac{-1}{36(a^{2}+b^{2})}\)
=x= \(\frac{1}{2}\)
And , \(\frac{-y}{12(a^{2}+b^{2})}=\frac{-1}{36(a^{2}+b^{2})}\)
=y=\(\frac{1}{3}\)
∴, the solution of the given pair of equations are x=\(\frac{1}{2}\) and y=\(\frac{1}{3}\) respectively.
25.
\(\frac{a^{2}}{x}-\frac{b^{2}}{y}=0\)
\(\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b\)
Solution:
Given equations are:
\(\frac{a^{2}}{x}-\frac{b^{2}}{y}=0\)
\(\frac{a^{2}b}{x}+\frac{b^{2}a}{y}=a+b\)
Taking \(\frac{1}{x}\) =u
Taking \(\frac{1}{y}\)=v
The pair of equations becomes:
a2u-b2v=0
a2bu+b2av-(a+b) =0
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= a2 , b1 = -b2 , c1=0
a2= a2b , b2 = b2a , c2=-(a+b)
By cross multiplication method
= \(\frac{u}{b^{2}(a+b)}=\frac{v}{a^{2}(a+b)}=\frac{1}{a^{2}b^{2}(a+b)}\)
Now, \(\frac{u}{b^{2}(a+b)}=\frac{1}{a^{2}b^{2}(a+b)}\)
=x= \(\frac{1}{a^{2}}\)
And, \(\frac{v}{a^{2}(a+b)}=\frac{1}{a^{2}b^{2}(a+b)}\)
=y= \(\frac{1}{b^{2}}\)
∴, the solution of the given pair of equations are x=\(\frac{1}{a^{2}}\) and y=\(\frac{1}{b^{2}}\) respectively.
26.
mx-my=m2+n2
x+y=2m
Solution:
Given equations are:
mx-my=m2+n2………………………………… (i)
x+y=2m…………………………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
a1= m , b1 = -n , c1=-(m2+n2)
a2= 1 , b2 = 1 , c2=-(2m)
By cross multiplication method
\(\frac{x}{(m+n)^{2}}=\frac{-y}{-m^{2}+n^{2}}=\frac{1}{m+n}\)
Now, \(\frac{x}{(m+n)^{2}}=\frac{1}{m+n}\)
=x= m+n
And, \(\frac{-y}{-m^{2}+n^{2}}=\frac{1}{m+n}\)
=y=m-n
∴, the solutions of the given pair of equations are x=m+n and y=m-n respectively.
27.
\(\frac{ax}{b}-\frac{by}{a}=a+b\)
ax-by=2ab
Solution:
Given equations are:
\(\frac{ax}{b}-\frac{by}{a}=a+b\) ……………………….. (i)
ax-by=2ab …………………………….. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
Here, a1= \(\frac{a}{b}\) , b1 = – \(\frac{b}{a}\) , c1=-(a+b)
Here, a2= a , b2 = – b, c2=-(2ab)
By cross multiplication method
= \(\frac{x}{b(b-a)}=\frac{-y}{a(-a+b)}=\frac{1}{b-a}\)
Now, \(\frac{x}{b(b-a)}=\frac{1}{b-a}\)
=x=b
And , \(\frac{-y}{a(-a+b)}=\frac{1}{b-a}\)
=y=-a
∴, the solution of the given equations are x=b and y=–a respectively.
28.
\(\frac{b}{a}x+\frac{a}{b}y-(a^{2}+b^{2})=0\)
x+y-2ab=0
Soln:
Given equations are:
\(\frac{b}{a}x+\frac{a}{b}y-(a^{2}+b^{2})=0\)…………………….. (i)
x+y-2ab=0…………………………………………. (ii)
The equations are of the form ax+by+c=0
We know that,
a1x+b1y+c1=0
a2x+b2y+c2=0
⇒\(\frac{x}{b_{1}c_{2}-b_{2}c_{1}}=\frac{-y}{a_{1}c_{2}-a_{2}c_{1}}=\frac{1}{a_{1}b_{2}-a_{2}b_{1}}\)
Here, According to the question,
Here, a1= \(\frac{b}{a}\) , b1 = Â \(\frac{a}{b}\) , c1=-(a2+b2)
Here, a2= 1 , b2 = -1, c2=-(2ab)
By cross multiplication method
= \(\frac{x}{b^{2}-a^{2}}=\frac{-y}{-b^{2}+a^{2}}=\frac{1}{\frac{b^{2}-a^{2}}{ab}}\)
Now, \(\frac{x}{b^{2}-a^{2}}=\frac{1}{\frac{b^{2}-a^{2}}{ab}}\)
=x=ab
And, \(\frac{-y}{-b^{2}+a^{2}}=\frac{1}{\frac{b^{2}-a^{2}}{ab}}\)
=y=ab
∴, the solutions of the given pair of equations are x=ab and y=ab respectively.