# If three times the larger of two digit numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If 5 times the smaller is divided by larger, we get 3 as quotient and 5 as remainder. Find the numbers.

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**Hint:**Above question is based on a linear equation in two variables.

1. Linear equation in two variables – An equation is said to be a linear equation in two variables if it is written in the form of $ax + by + c = 0,$ where a, b are real numbers and called as coefficients of x and y.

And c is a constant.

Here a and b should not equal 0.

2. The solution of linear equations in two variables – Solution of equation are the values of x and y. Which can satisfy the given equations.

3. When we draw these equations on graphs. We will get a graph of straight lines. And the number of solutions of a given equation depends upon the type of relation those linear functions have with each other.

Like if both the equations makes –

(i) Intersecting lines (unique solution) – When both lines intersect each other, they have a unique solution.

(ii) Parallel lines (no solution) – When both the lines never intersect each other, there is no solution for them.

(iii) Coincidence lines – When both the lines are overlapping, they have infinite no. of solutions.

4. We can find solutions from any of the following methods –

(i) Graphical method

(ii) Substitution method

(iii) Elimination method

(iv) Cross multiplication method

**Complete step by step solution:**Let the numbers be x and y.

Where x is larger than y.

Now according to the question we will form the equations.

It is given that,

1. If three times the larger of two digit numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder.

$Q = 4;$ $R = 8,$ $Divisor = y;$ $Dividend = 3x$

We know that –

$Dividend = (Divisor \times quotient) + remainder$

We will substitute the respective values in above relation

$3x = (y \times 4) + 8$

or $3x - 4y - 8 = 0$ equation (1)

2. If 5 times the smaller is divided by larger, we get 3 as quotient and 5 as remainder.

$Q = 3;$ $R = 5;$ $Divisor = x;$ $Dividend = 5y$

Substituting these values in above mentioned relation –

$5y = (x \times 3) + 5$

or $3x - 5y + 5 = 0$ equation (2)

These two equations are equations in two variables. Which can be solved by many methods. But here we will solve them by elimination method.

From equation (1) and (2)

First we will arrange these equations then we will subtract them to get the value of a variable.

$(3x - 4y - 8 = 0) - (3x - 5y + 5 = 0)$

Minus sign will change the signs of all the terms [of equation (2)]

$3x - 4y - 8 = 0$

$\dfrac{{ - 3x + 5y - 5 = 0}}{{0 + 1.y - 13 = 0}}$

or, $y = 13$

Place this value of $y = 13$ in equation (1)

Equation (1) is

$3x - 4y - 8 = 0$

As $Y = 13$

So, $3x = 8 + 4 \times 13$

$3x = 8 + 52$

$3x = 60$

or, $x = 20$

Therefore those two digits numbers are 13 and 20.

**Note:**We can use substitution method for above question

We have two equation

$3x - 4y - 8 = 0$ equation (1)

$3x - 5y + 5 = 0$ equation (2)

From equation (1)

$x = \dfrac{{8 + 4y}}{3}$ equation (3)

Substitute this value of x in equation 2

$3x - 5y + 5 = 0$

$3\left( {\dfrac{{8 + 4y}}{3}} \right) - 5y + 5 = 0$

$8 - y + 5 = 0$

$y = 13$

Place this value in of y in equation 3

$x = \dfrac{{8 + 4y}}{3}$

$x = \dfrac{{8 + 4(13)}}{3} = \dfrac{{60}}{3} = 20$

$(x = 20;y = 13)$