One integer is 3 more than another. Their product is 70. How do you find the integers?

Question

Eva Adamson · Accepted Answer

#(7,10)# or #(-7,-10)#

Let the integers be #x# and #y# (where let’s say #x > y#)

Conditions given are

#x = y + 3#

#xy = 70#

First equation can be written as

#x = 70/x + 3#

#x - 70/x = 3#

#(x^2 - 70)/x = 3#

#x^2 - 70 = 3x#

#x^2 - 3x - 70 = 0#

#x^2 + 10x - 7x - 70 = 0#

#x(x + 10) - 7(x + 10) = 0#

#(x -7)(x + 10) = 0#

#x = -7# or #x = 10#

If #x = -7#

Value of #y# from first equation is

#y = x - 3 = -7 - 3 = -10#

If #x = 10#

Value of #y# from first equation is

#y = x - 3 = 10 - 3 = 7#

Therefore, two sets of integers are possible: #(-7,-10)# and #(7, 10)#

Abigail Allen · Answer

undefined To find the two integers:

1. Set up a system of equations based on the given information.
2. Let one integer be represented by $x$ and the other by $x + 3$.
3. Write the equation: $x(x + 3) = 70$.
4. Expand and simplify the equation: $x^2 + 3x = 70$.
5. Rearrange the equation into standard quadratic form: $x^2 + 3x - 70 = 0$.
6. Factor the quadratic equation: $(x - 7)(x + 10) = 0$.
7. Solve for $x$: $x = 7$ or $x = -10$.
8. Since the other integer is 3 more than $x$, the integers are $7$ and $10$.