Solve : 7x-2y = 3
112.-311
Answers
Answer:
Given equations are 7x - 2y = 3 ….(1)
22x - 3y = 16 ….(2)
Here, a1=7,b1=-2 and c1=3 [from (1)]
a22=22,b2=-3 and c2=16 [from (2)]
Now, a1a2=722 and b1b2=-2-3=23
Since a1a2≠b1b2. Hence, the given system has a unique solution.
By cross multiplication method, we have

x(-2)×(-16)-(-3)×(-3)=y(-3)×22-(-16)×7=17×(-3)-22×(-2)
⇒x32-9=y-66+112=1-21+44
implies x23=y46=123
When x23=123⇒x=1
and y46=123⇒y=2
Hence, x=1y=2} is the required solution.
(ii) The given system of equations is 3x + y = 17 ....(1)
8x + 11y = 37 ....(2)
Here a1=3,b1=1 and c1=17 [from (1)]
a2=8,b2=11 and c2=37 [from (2)]
Now a1a2=38,b1b2=111
Since a1a2≠b1b2. Hence, the given system has a unique solution.
We can write the equations as
3x+y-17=0and8x+11y-37=0
By cross multiplication method, we have

x1×(-37)-11×(-17)=y(-17)×8-(-37)×3=13×11-8×1
implies x150=y-25=125 When x150=125⇒x=6
and y-25=125⇒y=-1
Hence, x=6y=-1} is the required solution.
(iii) The given system of equations is 2x + 5y - 17 = 0 ....(1)
5x + 3y - 14 = 0 ....(2)
Here, a1=2,b1=5 and c1=-17 [from (1)]
a2=5,b2=3 and c2=-14 [from (2)]
Now, a1a2=25,b1b2=53
Since a1a2≠b1b2. Hence, the given system of equations has a unique solution.
By cross multiplication method, we have

x5×(-14)-3×(-17)=y(-17)×5-(-14)×2=12×3-5×5
implies x-70+51=y-85+28=16-25
implies x-19=y-57=1-19
implies When x-19=1-19⇒x=1
and y-57=1-19⇒y=3
Hence, x=1y=3} is the required solution.
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ANSWER
Consider the given equations.
7x−2y=20 ……… (1)
11x+15y=−23 …….. (2)
From equation (1), we get
y=
2
7x−20
On putting value of y in equation (2), we get
11x+15(
2
7x−20
)=−23
11x+
2
105x
−150=−23
2
127x
=127
x=2
Therefore,
y=
2
7×2−20
y=−3
Hence, the value of x and y is 2,−3.
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