Question

5. Find two numbers such that the sum of twice the first and thrice the
second is 92, and four times the first exceeds seven times the second
by 2.​

Answer 1

Answer:

two numbers are such that the sum of twice the first and thrice the second is 92, and four times the first exceeds seven times the second by 2.

To find : the two numbers

Let x and y be the two numbers required.

According to the question :

=> 2x + 3y = 92 ...............................(1)

=> 4x- 7y = 2.....................................(2)

multiply the first equation by 2 , and subtract eq (1) from eq (2) , we get

4x + 6 y = 184

- (4x - 7y = 2)

---------------------

13 y = 182

-------------------------

=> y = 14 and x = 25 Answer

y = 14 and

Answer 2

Solution :

$\bf{\red{\underline{\bf{Given\::}}}}$

The sum of twice the first and thrice the second is 92 and four times the first exceeds seven times the second by 2.

$\bf{\red{\underline{\bf{To\:find\::}}}}$

The two number.

$\bf{\red{\underline{\bf{Explanation\::}}}}$

Let the two number be r and m.

A/q

$\bullet\sf{2r+3m=92..................(1)}\\\bullet\sf{4r-7m=2..................(2)}}$

$\underline{\underline{\bf{Using\:Substitution\:method\::}}}}}$

From equation (2),we get;

$\longrightarrow\sf{4r-7m=2}\\\\\longrightarrow\sf{4r=2+7m}\\\\\longrightarrow\sf{r=\dfrac{2+7m}{4} .........................(3)}$

Putting the value of r in equation (1),we get;

$\longrightarrow\sf{2\bigg(\dfrac{2+7m}{4} \bigg)+3m=92}\\\\\\\longrightarrow\sf{\dfrac{4+14m}{4} +3m=92}\\\\\\\longrightarrow\sf{4+14m+12m=368}\\\\\\\longrightarrow\sf{4+26m=368}\\\\\\\longrightarrow\sf{26m=368-4}\\\\\\\longrightarrow\sf{26m=364}\\\\\\\longrightarrow\sf{m=\cancel{\dfrac{364}{26} }}\\\\\\\longrightarrow\sf{\purple{m=14}}$

Putting the value of m in equation (3),we get;

$\longrightarrow\sf{r=\dfrac{2+7(14)}{4} }\\\\\\\longrightarrow\sf{r=\dfrac{2+98}{4} }\\\\\\\longrightarrow\sf{r=\cancel{\dfrac{100}{4} }}\\\\\\\longrightarrow\sf{\purple{r=25}}$

Thus;

The value of two number is r = 25 and m = 14 .