Math, asked by radhavig71, 1 year ago

In a class no of girls is x and that of boys is y also the no of girls is 10 more than the no of boys write the given data is in the form of linear equation in two variables also represent graphically find graphically the no of girls if the no of boys is 20?

Answers

Answered by fanbruhh
79
hey

here is answer


Question-

In a class no of girls is x and that of boys is y also the no of girls is 10 more than the no of boys write the given data is in the form of linear equation in two variables also represent graphically find graphically the no of girls if the no of boys is 20?


given -

In a class no of girls is x and that of boys is y also the no of girls is 10 more than the no of boys

and

no of boys=20

now

according to question


no of boys= y

no of girls=x

and

no of girls is. 10 more than that of boys


so

it will be in linear equation as

x= y+10

so

x-y=10............(1)

now

no of boys=20


put the value in equation (1)


x-20=10

x=10+20

x=30

hence no of girls=30


hope it helps

thanks

fanbruhh: thanks
deepti100428: we can also make equation such as x=y+10
deepti100428: and then solve it
Answered by Anonymous
77

\huge\mathfrak\red{Answer :}

According to question we have given that;

In a class number of girls = x

And number of boys = y

If girls (x) are 10 more than boys (y)

Then;

x - y = 10 .... (1)

or y = x - 10

We have to write the given date in tabular form...

Now, if x = 0

then,

y = 0 - 10

y = - 10

If x = 10 then;

y = 10 - 10

y = 0

If x = 5 then;

y = 5 - 10

y = - 5

And if x = -5 then;

y = - 5 - 10

y = - 15.

Now, the aquired data is...

x = 0, y = - 10
x = 10, y = 0
x = 5, y = - 5
x = -5, y = -15

[It's graph I in Attachment !!]

Also we have given that...

Number of boys = 20

So, number of girls.... = ?

Put the number of boys in eq. (1)

x - 20 = 10

x = 10 + 20

x = 30

\textbf{Number of boys = 20}

AND

\textbf{Number of girls = 30}

===================================



Attachments:

fanbruhh: perfect
parth340: not perfect
fanbruhh: welcome
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