Let's represent the two numbers by x and y. Then xy=60. The smaller number here is x=y-7.
Then (y-7)y=60, or y^2 - 7y - 60 = 0. Use the quadratic formula to (1) determine whether y has real values and (2) to determine those values if they are real:
discriminant = b^2 - 4ac; here the discriminant is (-7)^2 - 4(1)(-60) = 191. Because the discriminant is positive, this equation has two real, unequal roots, which are
-(-7) + sqrt(191)
y = -------------------------
-2(1)
and
-(-7) - sqrt(191)
y = ------------------------- = 3.41 (approximately)
-2(1)
Unfortunately, this doesn't make sense, since the LCM of two numbers is generally an integer.
Try thinking this way: If the LCM is 60, then xy = 60. What would happen if x=5 and y=12? Is xy = 60? Yes. Is 5 seven less than 12? Yes.
Answer:
all work pictured and shown
Answer:
we need to find out if the following get x = 2 as a final product
-2(x-4) = 4
-2x + 8 = 4
-2x = -4
x = 2
so it is a solution
26/x = 13
26 = 13x
x = 26/13
x = 2
so it’s a solution
-3.8x = -7.4
x = 7.4/3.8 ≠ 2
you get ≈ 1.95
so it’s not a solution
4(x-1) - 3(x-2) = -8
4x - 4 - 3x + 6 = -8
x + 2 = -8
x = -10
so it’s not a solution
The answer is
X=7
Have a nice day <3
