Question

Seven years ago,a father was seven times as old as his daughter. Three years from now, he will be three times as old as his daughter will be . If present ages of father and daughter are x and y respectively, represent this situation algebraically and graphically

Answer 1

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Answer 2

$<b><huge>Answer<b><huge>$

Let to be Present age of father =x

And Present age of his Daughter =Y

*
$\huge{ \fbox{ \mathtt{seven \: years \: ago}}}$

Age of father =) x-7

Age of His Daughter=) y-7

Age of father=7× Age of his Daughter

7(x-7)=(y-7)

7x-49=y-7

7x-y=-7+49

7x-y=42 ( Equation-1)

$\huge{ \fbox{ \mathsf{according \: to \: the \: question}}}$

THREE YEARS HENCE,

AGE OF FATHER=X+3

AGE OF HIS DAUGHTER=Y+3

So.

Age of father=3×Age of his Daughter

3(x+3)=y+3

3x+9=y+3

3x-y=3-9

3x-y=-6.

The given situation can be represented algebraically by the system of simultaneous equation given by 7x – y = 42 and 3x – y = 6.

Consider the equation

7x – y = 42.

∴ y = 7x – 42

When x = 6,
 y = 7 × 6 – 42 = 42 – 42 = 0

When x = 7, 
y = 7 × 7 – 42 = 49 – 42 = 7

When x = 5, 
y = 7 × 5 – 42 = 35 – 42 = – 7

These points can be represented in a table :-

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