Question

A teacher is 18 years elder to her student .3 years ago one-tenth of the teacher's age was
equal to one-fourth of the student's age. Find
their present ages.​

Answer 1

Answer:

present ages is 28 teachers age and student's age is 7

Answer 2

$\huge\purple{Required \: Answer :}$

Let,

The Age of Teacher Be $x$ ,

The Age of Students Be $y$ .

From Condition 1,

$x = y + 18 \\ x - y = 18$

From Condition 2,

$\frac{1}{10} \times (x - 3) = \frac{1}{4} \times (y - 3) \\ \frac{x}{10} - \frac{3}{10} = \frac{y}{4} - \frac{3}{4} \\ \frac{x - 3}{10} = \frac{y - 3}{4}$

Multiply Each Term By 20 ,

$\frac{x - 3}{10} \times 20 = \frac{y - 3}{4} \times 20 \\ 2(x - 3) = 5(y - 3) \\ 2x - 6 = 5y - 15 \\ 2x - 5y = - 15 + 6 \\ 2x - 5y = - 9$

Multiply Equation 1 By 2,

$2x - 2y = 36$

Subtract Equation 2 From Equation 3 ,

$\: \: \: \: \: \: 2x - 2y = 36 \\ - \: \: \: \: 2x - 5y = - 9 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 3y = 45 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{45}{3} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = 15$

Put y = 15 in Equation 1,

$x - y = 18 \: \: \\ x - 15 = 18 \\ x = 18 + 15 \\ x = 33 \: \: \: \: \: \: \: \: \:$

Hence,

The Age Of Teacher is $\huge{33}$ years,

The Age Of Student is $\huge{15}$ years.

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