Question

one number is three more than a second number. the sum of their square is 37 more than the product of the numbers. find the two numbers.​

Answer 1

The two numbers are -7 and 4.

Step-by-step explanation:

• To solve this problem, we'd adopt a few steps.

Let $x$ be the second number and $x+3$ be the first number.

• It has been given that the sum of squares is $37$ more than the product of the numbers.

∴ $x^{2}+(x+3)^{2}=x(x+3)+37$

∴ $x^{2}+x^{2}+6x+9=x^{2}+3x+37$

We'll set the equation to zero by subtracting $x^{2}+3x+37$ from each side:

∴ $(2x^{2}+6x+9)-(x^{2}-3x-37)=0$

∴ $x^{2}+3x-28=0$

We'll factorize $x^{2}+3x-28$, with its factors being: $(x+7)(x-4)=0$

There will be two solutions to $(x+7)$ and $(x-4)$

1. $x=-7$
2. $x=4$

Hence, the two numbers that we require are -7 and 4.

Answer 2

Answer:

a)  x = -4 ; y = -7 (or)

b)  x = 7 ; y = 4

Step-by-step explanation:

let x and y be the numbers,

given ,

x = y+3   ------- (1)

x² + y² = xy + 37   ---- (2)

=> (y+3)² + (y)² = (y+3)y + 37       by (1),

=>  y² + 6y + 9 + y² = y² + 3y + 37

=> 2y² - y² + 6y - 3y + 9 - 37 = 0

=> y² + 3y - 28 = 0

=> y² + 7y - 4y - 28 = 0

=> y(y+7) -4(y+7) = 0

=> (y+7)(y-4) = 0

=> y + 7 = 0    or y-4 = 0

=> y = -7      or y = 4

when y = -7,

x = -7+3 = -4

so, one of the solutions is (-4,-7)

when y = 4,

x = 4+3 = 7

so, the other solution is (7,4)