1) f(x)=2x+6
f(2)=2(2)+6
=4+6
=10
2)f(x)=3x
f(a+1)=3(a+1)
=3a+3
3)f(x)=3x-1 and g(x)=5x+3
f(2)=3(2)-1
f(2)=6-1
=5
g(3)=5x+3
=5(3)+32
=15+3
=18
f(2)+f(3)=5+18
=23.
1. 5^2 = 25
2. 2^6 = 64
3. 25^(1/2) =5
A b + a c - 4 b - 4 c = a ( b + c ) - 4 ( b + c ) =
= (b + c) (a - 4)
You do 3/5 minus 3/5 which cancels out to 0. Then you do 6 minus 4 which is 2. The answer is 2.
Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as <span>y=(3/4)X+5
Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:
4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)
Let 's now solve the system
4y=3x + 20
4x-3y=-1
We are to solve this system using subtraction. To accomplish this, multiply the first equation by 3 and the second equation by 4. Here's what happens:
12y = 9x + 60 (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)
Then we have
12y = 9x + 60
-12y =-16x - 4
If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x: 7x = 56; x=8
We were given equations
</span><span>y=3/4X+5
4x-3y=-1
We can subst. x=8 into either of these eqn's to find y. Let's try the first one:
y = (3/4)(8)+5 = 6+5=11
Then x=8 and y=11.
You should check this result. Subst. x=8 and y=11 into the second given equation. Is this equation now true?</span>
