Answer: See below
Explanation:
6a + 11 = 2a + 83
6a - 2a = 83 - 11
4a = 72
a = 72/4
a = 18
Answer:
x<15
Step-by-step explanation:
x needs to be less than 15 since Courtney's not paying any more than that.
If we can match teh bases we can solve
because if x=x and xᵃ=xᵇ, we can conclude that a=b
16=2⁴
32=2⁵
rememeber that




2=2 so we conclude that 4(3x+2)=5(-2x-7)
4(3x+2)=5(-2x-7)
expand/distribute
12x+8=-10x-35
add 10x both sides
22x+8=-35
minus 8 both sides
22x=-43
divide both sides by 22
x=-43/22
There are
ways of picking 2 of the 10 available positions for a 0. 8 positions remain.
There are
ways of picking 3 of the 8 available positions for a 1. 5 positions remain, but we're filling all of them with 2s, and there's
way of doing that.
So we have

The last expression has a more compact form in terms of the so-called multinomial coefficient,

<h3><u>The value of the greater number is 15.</u></h3><h3><u>The value of the smaller number is 7.</u></h3>
x = 2y + 1
3x = 5y + 10
Because we have a value for x we can plug this value in to find the value of y.
3(2y + 1) = 5y + 10
Distributive property.
6y + 3 = 5y + 10
Subtract 5y from both sides.
y + 3 = 10
Subtract 3 from both sides.
y = 7
We can plug this value back into the original equation to find the value of x.
x = 2(7) + 1
x = 15