Question

The taxi fare in a city is charged as per the rates stated below: Rate for the first kilometre of journey is Rs.5 and the rate for the subsequent distance covered is Rs 4 per km. Taking distance covered as x km and total fare as Rs y, the linear equation in two variables is framed as

Answer 1

$\large\bf\underline{Given:-}$

• Rate for first kilometre = Rs 5
• the rate for the subsequent distance covered is Rs 4 per km.

$\large\bf\underline {To \: find:-}$

• Linear equation

$\huge\bf\underline{Solution:-}$

• total fare = Rs y
• Total distance covered = x km
• Fare for 1st km = Rs 5
• Subsequent distance = 4/km

Fare for rest distance = 4(x - 1)

Total fare = fare for 1st km + fare for rest distance

»» y = 5 + 4(x-1)

»» y = 5 + 4x - 4

»» y = 4x + 1

So ,the Linear equation formed is » y = 4x + 1

Answer 2

✯✯ QUESTION ✯✯

The taxi fare in a city is charged as per the rates stated below: Rate for the first kilometre of journey is Rs.5 and the rate for the subsequent distance covered is Rs 4 per km. Taking distance covered as x km and total fare as Rs y, the linear equation in two variables is framed as

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✰✰ ANSWER ✰✰

$\longmapsto\tt{Let\:the\:distance\:covered\:be=x\:km}$

$\longmapsto\tt{Fare\:for\:travelling=y\:Rs}$

$\longmapsto\tt{Fare\:for\:1st\:Km=5\:Rs.}$

$\longmapsto\tt{Fare\:for\:Subsequent\:km=5\:Rs}$

➥Now ,

$\longmapsto\tt{y=5\times{1}+(x-1)\times{4}}$

$\longmapsto\tt{y=5+4x-4}$

$\longmapsto\tt{y=4x+5-4}$

$\longmapsto\tt{y=4x+1}$

☛So , The linear equation is y = 4x + 1 ....