Linear Equation In Two Variables
Question 1. Solve for x and y:
7+y=17,
7-4=1
Solution: Given
x+y=17 → (1)
x-4=1 → (2)
From equation 1 we get y= 17-x → R
Substituting the value of from equation in 2, we get
x-(17-x)=1
2-17+2=1
2x=1+17
2x = 18
x = \(\frac{18}{2}\) ⇒ x=9
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Substituting the value of x in an equation, we
y= 17-9
y= 8
Solution is x=9
y=8
2. x+2y=19
-x+2y=1
Solution: Given x+2y= 190 → (1)
-x+2y=1 → (2)
From equation 1 we get x=19-2y → 3
Substituting the value of and from equations (3)and (2) we get
-(19-2y) +2y= 1
– 19+2y+24=1
4y = 1+19
4y = 20
y = \(\frac{20}{4}\)
y = 5
Substitute value of y in equation 3 we get
x=19-25
X=19-2(5)
1=19-10
x=9
Solution is x =9
y=5
3. x – y= 0.9
\(\frac{11}{2(x+y)}=1\)Solution: Given x – y= 0.9 → (1)
⇒ \(\frac{11}{2(x+y)}=1\)
⇒ \(\frac{11}{2 x+2 y}=1\)
2x + 2y = 11 → (2)
From equation we get x = 0-9 + y → (3)
Substituting the value of x from the equation, we get
2(0.9+4)+2y=11
1.8 +25+24=11
4y = 11-1-1
4y= 9, 2
y = \(\frac{9.2}{4} \Rightarrow y=2.3\)
Substituting the value of the y value in the (3) equation we get
2=0·9+2·3
X = 3.2
Solution is 2=3.2
y=2-3
4. 3x-2y=6
\(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\)Solution: Given 3x-2y=6
⇒ \(\frac{x}{3}-\frac{y}{6}=\frac{1}{2}\)
⇒ \(\frac{6 x-y}{6}=\frac{1}{2}\)
2(6x-4)=6
12x-2y=6
From equation 1 we get 3x-2y=6
3x=6+24
⇒ \(x=\frac{2 y+6}{3} \rightarrow \text { (3) }\)
Substituting the value of x from equation 3 in 2, we get
⇒ \(12\left(\frac{2 y+6}{3}\right)-2 y=6\)
8y+824-2y=6
6y=6-24
6y=-18
⇒ \(y=\frac{-18}{6} \Rightarrow y=-3\)
Substituting the Value of y in Equation 3
⇒ \(x=\frac{2(-3)+6}{3}\)
⇒ \(x=\frac{-6+6}{3} \Rightarrow x=0\)
Solution is x=0
y=-3
5. 0.4x+0.34 = 1.7
07x-0-2y=0·8
Solution: Given 0.4x+0-3y=1-7 → (1)
0.72-0·24=0·8 → (2)
From equation (1) we get 0.4x=1.7-0.34
⇒ \(x=\frac{1.7-0.3 y}{0.4} \rightarrow \text { (3) }\)
Substituting the value of x from equation (3) in (2)we get
⇒ \(0.7\left(\frac{1.7-0.3 y}{0.4}\right)-0.2 y=0.8\)
⇒ \(\frac{1.19}{0.4}-\frac{0.21 y}{0.4}-0.2 y=0.8\)
2.975-0.525y-0-2y=0.8
-0-725y = 0. = 0 .525y-2.975
– 0.725y = 2175
⇒ \(0.7\left(\frac{1.7-0.3 y}{0.4}\right)-0.2 y=0.8\)
⇒ \(\frac{1.19}{0.4}-\frac{0.21 y}{0.4}-0.2 y=0.8\)
2.975y – 0.525y – 0.2y = 0.8
-0.725y = 0.8 – 2.975
-0.725y = -2.175
\(y=\frac{-2.175}{-0.725}\)Substituting the value of ‘y’ from the equation
⇒ \(x=\frac{1.7-0.3(3)}{0.4}\)
⇒ \(x=\frac{1.7-0.9}{0.4}\)
⇒ \(x=\frac{0.8}{0.4}\)
x = 2
The solution is
y =3
x =2
6. y = 2x – 6
y= 0
Solution:
Given equations are
y = 2x-6 (1)
y= 0 (2)
Substituting y value in Equation (1)
0=2x-6
-2x = -6
⇒ \(x=\frac{6}{2} \Rightarrow x=3\)
Solution is x = 3
y = 0
7. 65x -33y = 97
33x-65y = 1
Solution: Given equations are 65x-33y=97 (1)
33x-65y = 10 (2)
on adding equation ( & we get
⇒ 98x-98y=98 x-y=1 → (3)
On Subtracting equation (2)from (1) we get
– 32x-32y=-96
– 32(x+y)=+96
(x + Y) = \(\frac{96}{32}\)
x + y = 3 → (4)
Adding equation (3) and (4)
putting x = 2 in equation (3)
2-y=1 =) -y= 1-2 =) -y=-1 =) y=1
Hence, the Solution is x=2
4 = 1
8. 217x+13ly=913
131x+217y=827
Solution: Given equations 217x+131y=9130 (1)
131x +2174=827 (2)
on adding (1) and (2) we get
348x+3484 = 1740
348(x+y)=1740
⇒ \(x+y=\frac{1740}{348}\) → (3)
On Subtracting the equation we get
-86x+86y=-86
-86(x-4)=-86
x-y=1 → (4)
Adding equation (3) and (4)
2x = 6
⇒ \(x=\frac{6}{2} \Rightarrow x=3\)
putting x = 3 in equation (3)
3 + y = 5
y = 5 – 3
y = 2
Hence the solution is x = 3
y = 2
Question 2. Solve the following System of equations by using the cross-multiplication method:
1. 2x+y=5
3x+2y=8
Solution: The given equation can be written as
2x+y-5=0
3x+2y-8=0
By Gross multiplication method, we get
x y 1
1 -5 2 1
2 -8 3 1
⇒ \(\frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3}\)
⇒ \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1}\)
⇒ \(\frac{x}{2}=\frac{1}{1} \text { or } x=2\)
and \(\frac{y}{1}=\frac{1}{1} \text { or } y=1\)
Hence, x=2 and y=1 is the required Solution.
2. 8x+13y-29=0
12x-74-17=0
Solution:
The given equations are
8x+13y-29=0
12x-7y-17=0
By Cross multiplication method, we get
x y 1
13 -29 8 13
-7 -17 12 -7
⇒ \(\frac{x}{-221-203}=\frac{y}{-348+136}=\frac{1}{-56-156}\)
⇒ \(\frac{x}{-424}=\frac{y}{-212}=\frac{1}{-212}\)
⇒ \(\frac{x}{-424}=\frac{1}{-212}\)
⇒ \(x=\frac{-424}{-212}\)
x = 2
and \(\frac{y}{-212}=\frac{1}{-212}\)
y = 1
Hence, x=2 and y=1 is the required Solution.
3. 2y+2x=0
4y+3x=5
Solution: The given equation Can be written as 2x +3y=0 3x+4y-5
By Cross multiplication method, we get
x y 1
3 0 2 3
4 -5 3 4
⇒ \(\frac{x}{-15-0}=\frac{y}{0+10}=\frac{1}{8-9}\)
⇒ \(\frac{x}{-15}=\frac{y}{10}=\frac{1}{-1}\)
when \(\frac{x}{-15}=\frac{1}{-1} \Rightarrow x=15\)
and\(\frac{y}{10}=\frac{1}{-1} \Rightarrow y=-10\)
Hence, x=15 and y=-10 is the required solution.
4. \(\frac{x}{6}+\frac{4}{15}=4\)
\(\frac{x}{3}-\frac{4}{12}=\frac{19}{4}\)Solution: The given equation can be written as
⇒ \(\frac{x}{6}+\frac{y}{15}-4=0\)
⇒ \(\frac{5 x+2 y-120}{30}=0\)
5 x+2 y-120 = 0 → (1)
⇒ \(\frac{x}{3}-\frac{y}{12}-\frac{19}{4}=0\)
⇒ \(\frac{8 x-2 y-114}{24}=0\)
8x – 2y – 114 = 0→ (2)
By Cross multiplication method & we get
x y 1
2 – 120 5 2
-2 -114 8 2
⇒ \(\frac{x}{-225-240}=\frac{y}{-960+570}=\frac{1}{-10-16}\)
⇒ \(\frac{x}{-468}=-\frac{y}{-390}=\frac{1}{-26}\)
when \(\frac{x}{-4 6 8}=\frac{1}{-26}\)
⇒ \(x=\frac{-468}{-26} \Rightarrow x=18\)
⇒ \(\text { and } \frac{y y}{-390}=\frac{1}{-26} \Rightarrow y=\frac{-390}{- 20} \Rightarrow y=15\)
Hence X-18 and y = 15 is the required solution.
5. x+y=a+b
ax-by=a2=62
Solution: The given equations Can be written as
x+y=(a-6)=0
ax-by-(a2+63)=0
By Cross multiplication method, we get
x y 1
1 -(a-b) 1 1
-b -(a2+b2) a -b
⇒ \(\frac{x}{-\left(a^2+b^2\right)-b(a-b)}=\frac{y}{-a(a-b)+\left(a^2+b^2\right)}=\frac{1}{-b-a}\)
⇒ \(\frac{x}{-a^2-b^2-a b+b^2}=\frac{y}{-a^2+a b+a^2+b^2}=\frac{1}{-b-a}\)
⇒ \(\frac{x}{-a^2-a b}=\frac{y}{a b+b^2}=\frac{1}{-b-a}\)
when \(\frac{x}{-a^2-a b}=\frac{1}{-b-a}\)
⇒ \(x=\frac{-b(a+b)}{-(a+b)}\)
x = a
and \(\frac{y}{a b+b^2}= \frac{1}{-b-a}\)
⇒ \(\frac{y}{-b(-a-b)}=\frac{1}{-b-a}\)
⇒ \(y=\frac{-b(-b-a)}{+b-a}\)
Hence x=a and y = b are the required Solution
6. ax-by=a2+b2
x+y=20
Solution: The given equations Can be written as,
ax-by-(a2=+b2)=0
x+y-2a=0
By Cross multiplication method, we get
x y 1
-b -(a2+b2) a -b
1 -2a 1 1
⇒ \(\frac{x}{2 a b+\left(a^2+b^2\right)}=\frac{y}{-\left(a^2+b^2\right)+2 a^2}=\frac{1}{a+b}\)
⇒ \(\frac{x}{2 a b+\left(a^2+b^2\right)}=\frac{y}{-\left(a^2+b^2\right)+2 a^2}=\frac{1}{a+b}\)
⇒ \(\frac{x}{(a+b)^2}=\frac{y}{a^2-b^2}=\frac{1}{a+b}\)
⇒ \(\frac{x}{(a+b)^2}=\frac{1}{a+b}\)
⇒ \(x=\frac{(a+b)^2}{(a+b)}\)
x = a+b
and \(\frac{y^2}{a^2-b^2}=\frac{1}{a+b}\)
⇒ \(y=\frac{(a+b)(a-b)}{(a+b)}\)
y = a-b
Hence x= a+b an is the required solution.
7. ax+by = c
bx+ay=1+c
Solution: The given equations can be written as
ax+by = c
bx+ay= (C+1)=0
B Cross multiplication method, we get
x y 1
b -c a b
a -(c+1) b a
⇒ \(\Rightarrow \frac{x}{-b(c+1)+a c}=\frac{y}{-b c+a(c+1)}=\frac{1}{a^2-b^2}\)
⇒ \(\frac{x}{-b c-b+a c}=\frac{y}{-b c+a c+a}=\frac{1}{a^2-b^2}\)
⇒ \(x=\frac{ac-b-b c}{a^2-b^2}\)
and \(\frac{y}{-b c+a c+a}=\frac{1}{a^2-b^2}\)
⇒ \(y=\frac{-(b c-a-a c)}{-\left(-b^2-a^2\right)} \Rightarrow y=\frac{b c-a-a c}{b^2-a^2}\)
Hence x \(=\frac{a c-b-b c}{a^2-b^2} \text { and } y=\frac{b c-a-a c}{b^2-a^2}\).
8. (a-b)x + (a+b)y = a2- 2ab-b2
(a+b) (x+y)= a2+b2
Solution:
The given equation can be written as
(a-b)x + (a+b)y – (a2+2a+b2) = 0
(a-b)x + (a+b)y – (a+b)2 = 0 (1)
(a+b)(x+y) = a2+b2
ax+ay+bx+by – (a2-b2) = 0
(a+b)x + (a+b)y – (a-b)2 = 0 (2)
By cross-multiplication methods (1)and(2)
x y 1
(a+b) -(a+b)2 (a-b) (a+b)
(a+b) -(a2-b2) (a+b) (a+b)
⇒ \(\frac{x}{-(a+b)\left(a^2-b^2\right)+(a+b)^3}=\frac{y}{-(a+b)^3+\left(a^2-b^2\right)(a-b)}=\frac{1}{(a-b)(a+b)-(a+b)^2}\)
⇒ \(\frac{x}{-a^3+a b^2-a^2 b+b^3+a^3+b^3+3 a^2 b+3 a b^2}\)
⇒ \(=\frac{y}{-p^2-p^2-3 a^2 b-3 a b^2+a^3-a^2 b-a b^2+b^3}\)
⇒ \(=\frac{1}{a\not^2-b^2-\not a^2-b^2-2 a b}\)
⇒ \(\frac{x}{2 b^3+4 a b^2+2 a^2 b}=\frac{y}{-4 a^2 b-4 a b^2}=\frac{1}{-2 b^2-2 a b}\)
⇒ \(\frac{x}{2\left(b^3+2 a b^2+a^2 b\right)}=\frac{y}{-4\left(a^2 b-a b^2\right)}=\frac{1}{-2 b(b+a)}\)
when \(\frac{x}{2\left(b^3+2 a b^2+1 a^2 b\right)}=\frac{1}{-2 b(b+a)}\)
⇒ \(x=\frac{2\not\left(b^3+2 a b^2+a^2 b\right)}{-2 b(b+a)}\)
⇒ \(x=\frac{\not b\left(b^2+2 a b+a^2\right)}{\not b(b+a)}\)
⇒ \(x=\frac{(a+b)^2}{(a+b)}\)
x = a+b
and\(\frac{y}{-4\left(a^2 b-a b^2\right)}=\frac{1}{-2 b(b+a)}\)
⇒ \(y=\frac{-4 b\left(d^2-a b\right)}{-2 b(b+a)}\)
⇒ \(y=\frac{2 a^2-2 a b}{a+b} \times \frac{1}{2 a^2}\)
⇒ \(y=\frac{-2 a b}{a+b}\)
Hence x=a+b and \(y=\frac{-2 a b}{a+b}\) is the required solution.
Question 3. Each of the following System of equations determines whether the System has a unique solution, no Solution, or infinitely many Solutions. In Case there is a unique solution, find it:
1. 2x-3y=17
4x+9=13
Solution: Given equations are
2x-3y=170 (1)
4x+y=13 (2)
Here a1 =2, b1 =-3, C1 = 17
a2 =4, b2 =1 and C2 =13
Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2} \text { and } \frac{b_1}{b_2}=\frac{-3}{1}\)
Since\(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} .\) Hence, the given system has a unique solution.
By Cross multiplication method, we have
x y 1
3 -17 2 -3
1 -13 4 1
⇒ \(\frac{x}{39+17}=\frac{y}{-68+26}=\frac{1}{2+12}\)
⇒ \(\frac{x}{56}=\frac{y}{-42}=\frac{1}{14}\)
When \(\frac{x}{56}=\frac{1}{14}\)
⇒ \(x=\frac{56}{14} \Rightarrow x=4\)
And \(\frac{y}{-42}=\frac{1}{14}\)
⇒ \(y=\frac{-42}{14} \Rightarrow y=-3\)
Hence x=4 and y=-3 is the required solution.
2. 5x+2y=16
3x+ \(\frac{6}{5}\) y = 2
Solution: Given equation are 5x+2y=16
3x+\(\frac{6}{5}\)
Here a1=5, b1=2, c1=16
a2=3, b2 = \(\frac{6}{5}\), c2=2
Now,
⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)
⇒ \(\frac{5}{3}=\frac{2}{6 / 5} \neq \frac{16}{2}\)
⇒ \(\frac{5}{3}=\frac{10}{6} \neq 8\)
⇒ \(\frac{5}{3}=\frac{5}{3} \neq 8\)
Since,\(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2} .\) hence the given system has no solution.
3. 3x+4=2
6x+2y=4
Solution: Given equations are 3x+y=2
6x+2y=4
Here a1 = 3, b1 = 1, c1=2
a2=6, b2=2, c2=4
Now,
⇒ \(\frac{a_1}{a_2}=\frac{3}{6} \text { and } \frac{b_1}{b_2}=\frac{1}{2} \text { and } \frac{c_1}{c_2}=\frac{2}{4}\)
Since \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} .\) Hence the given system has an infinite solution.
4. \(\frac{x}{3}+\frac{y}{2}=3\)
x-2y = 2
Solution: Given equations are \(\frac{x}{3}+\frac{y}{2}=3\)
x-2y = 2
Here a1 = \(\frac{1}{3}\), b1 = \(=\frac{1}{2}\)
, c1=2
a2=6, b2=2, c2=4
Now
⇒ \(\frac{a_1}{a_2}=\frac{1 / 3}{1}=\frac{1}{3} \text { and } \frac{b_1}{b_2}=\frac{1 / 2}{-2}=\frac{-1}{4}\)
Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} .\) Here the given system has unique solutions
By Cross multiplication method, we get
x y 1
1/2 -3 1/3 1/2
-2 -2 1 -0
⇒ \(\frac{x}{-\left(\frac{1}{2}\right)(2)-6}=\frac{y}{-3+2\left(\frac{1}{3}\right)}=\frac{1}{-2\left(\frac{1}{3}\right)-1\left(\frac{1}{2}\right)}\)
⇒ \(\frac{x}{-1-6}=\frac{y}{-3+\frac{2}{3}}=\frac{1}{-\frac{2}{3}-\frac{1}{2}}\)
⇒ \(\frac{x}{-7}=\frac{y}{\frac{-9+2}{3}}=\frac{y}{\frac{-4-3}{6}}\)
⇒ \(\frac{x}{-7}=\frac{y}{-7 / 3}=\frac{1}{-7 / 6}\)
⇒ \(\frac{x}{-7}=\frac{-3 y}{7}=\frac{-6}{7}\)
⇒ \(\frac{x}{-7}=\frac{-3 y}{7}=\frac{-6}{7}\)
⇒ \(\text { when } \frac{x}{-7}=\frac{-6}{7}\)
⇒ \(x=\frac{-6(-7)}{7}\)
⇒ \(x=\frac{42}{7}\)
x = 6
and \(\frac{-3 y}{7}=\frac{-6}{7}\)
⇒ \(-3 y=\frac{-6(7)}{7}\)
7(-3)=-6(7)
-21y = -42y
⇒ \(y=\frac{42}{21}\)
y = 2
Hence x = 6 and y=2 is a required solution.
5. Kx+2y=5
3x+9=1
Solution: The given system of equations is
kx+2y=5
3x+y=1
Here, a1=k, b1 =2 and c1=5
a2=3, b2=1 and c2=1
The System has unique Solution \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
⇒ \(\frac{k}{3} \neq \frac{2}{1}\)
⇒ \(\Rightarrow \quad k \neq 6\)
So, k can take any real value except 6.
6. 4x+ky +1=0
2x+2y+2=0
Solution: The given System of equations is 4x+ky+8=0
2x+2y+2=0
Here a1 = 4, b1 = k and c1 =8
a2=2, b2 =2 and C2 =2
The System has unique solution if \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
⇒ \(\begin{aligned}
& \Rightarrow \frac{2 y}{x} \neq \frac{k}{2} \\
& \Rightarrow k \neq 4
\end{aligned}\)
So k can take any real value except 4.
7. 2x-37-5=0
kx-by-8=0
Solution: The given System of equations are 2x-34-5=0
kx-6x-8=0
Here, a1 =2, b1=3 and c1 =-5
a2=k, b2=6 and C2 =-8
The System has unique solution if \(\text { if } \frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)
⇒ \(\frac{2}{k}+\frac{-3}{-6}\)
⇒ \(\frac{2}{k}+\frac{1}{2}\)
⇒ \(\begin{aligned}
& \frac{4}{k} \neq 1 \\
&\quad k \neq 4
\end{aligned}\)
So k can take any real value except 4.
8. 8x+5y=9
kx+10y=18
Solution: The given System of equations is 8x+5y=9
Here a1 =8, b1=5 and c1 =9
a2 =k, b2 = 10 and c2 = 18
kx+10y=18
The System has infinitely many solutions if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
⇒ \(\frac{8}{k}=\frac{5}{10}=\frac{9}{18}\)
⇒ \(\frac{8}{k}=\frac{1}{2}=\frac{1}{2}\)
when \(\frac{8}{k}=\frac{1}{2} \Rightarrow k=16\)
Hence, for k = 16, the given System equations will have infinitely many solutions.
Question 4. The Sum of the two numbers is 85. Suppose the langer number exceeds four times the Smaller one bys. Find the numbers.
Solution: let the two numbers x and y, and x>y.
According to question x+y=85 → (1)
and X-4y=5 → (2)
Subtracting equation (2) from (1), we get
5y=80 =) y=16
Putting y=16 in equation (, we get
x+16=85
2=85-16
– x = 69