Question
Find the solution of the given system of equations.
x+y−82=x+2y−148=3x+y−1211
x+y−82=x+2y−148=3x+y−1211
Answer: Option B
:
B
Take first two components,
x+y−82=x+2y−148
⇒8(x+y−8)=2(x+2y−14)
⇒8x+8y−64=2x+4y−28
⇒6x+4y−36=0
⇒3x+2y−18=0.....(i)
Take last two components,
x+2y−148=3x+y−1211
⇒11(x+2y−14)=8(3x+y−12)
⇒11x+22y−154=24x+8y−96
⇒−13x+14y−58=0.....(ii)
On multiplying equation (i) by 7, we get
⇒21x+14y−126=0...(iii),
On subtracting equation (ii) from (iii), we get
⇒34x=68
⇒x=2
On substituting value of x=2 in equation (i), we get
3×2+2y−18=0
⇒y=6
∴ The solutionis (2, 6).
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:
B
Take first two components,
x+y−82=x+2y−148
⇒8(x+y−8)=2(x+2y−14)
⇒8x+8y−64=2x+4y−28
⇒6x+4y−36=0
⇒3x+2y−18=0.....(i)
Take last two components,
x+2y−148=3x+y−1211
⇒11(x+2y−14)=8(3x+y−12)
⇒11x+22y−154=24x+8y−96
⇒−13x+14y−58=0.....(ii)
On multiplying equation (i) by 7, we get
⇒21x+14y−126=0...(iii),
On subtracting equation (ii) from (iii), we get
⇒34x=68
⇒x=2
On substituting value of x=2 in equation (i), we get
3×2+2y−18=0
⇒y=6
∴ The solutionis (2, 6).
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