Question

find 2 no. such that five times the greater exceeds four time the lesser by 22 and three times the greater together with seven times lessseris 32 solve it on a paper and send to me

Answer 1

Answer: two numbers are 2 and 6.

Answer 2

Answer:

The two numbers x and y are 6 and 2 respectively.

Step-by-step explanation:

Let the two numbers be x and y, and x is greater than y.

From the first statement given, five times the greater exceeds four times the lesser by 22, can be written as

$5x=4y+22\implies 5x-4y=22$                         ...(1)

From the second statement given, three times the greater together with seven times lesser is 32, can be written as  

$3x+7y=32$                                                       ...(2)

To solve the algebraic expressions, multiply equation (1) with 3

$3(5x-4y=22)=15x-12y-66=0$                      ...(3)

Multiply equation (2) with 5

$5(3x+7y=32)=15x+35y-160=0$                    ...(4)

Substracting equation (3) from equation (4)

$15x+35y-160-(15x-12y-66)=0$

$15x+35y-160-15x+12y+66=0$

$47y-94=0$

$47y=94\implies y=\dfrac{94}{47}=2$

Using y = 2, in equation (1),

$5x-4\times 2=22$

$\implies 5x-8=22\implies 5x=22+8$

$\implies 5x=30\implies x=\dfrac{30}{5} =6$

Therefore, the two numbers x and y are 6 and 2 respectively.

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