Find four consecutive integers such that three times the sum of the first two integers exceeds the sum of the last two by 70.

we know the difference of consecutive integers is 1 so we can achieve every next integer by adding 1 in the previous integer .

let us start with assuming the four consecutive integers as :

x , x+1 , x+2 , x+3

3 times ( sum of first two ) = sum of last two + 70

3 ( x+ x+1) = x+2+x+3 + 70

3( 2x +1 ) = 2x + 75

6x + 3 = 2x + 75

6x-2x = 75 - 3

4x = 72

x= 72/4 = 18

so first integer is x= 18 next is 19 next 20 and next 21

Answer : 18 ,19,20,21

use substitution to solve each system of equations. y = x + 5 3x + y = 25 62/87,21 y = x + 5 3x + y = 25 substitute x + 5 for y in the second equation. substitute the solution for x into either equation to find y. the solution is (5, 10)

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step-by-step explanation:

hey! so 1 (25) out of 4 (100) is 25 so your answer is 25%. it's like quarters!