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>
A= 30
Explanation: 88-58=30
Answer:
1. 8x^3+6x^2+7
Step-by-step explanation:
Well I'm assuming that you want me to simplify the fraction 40/6. Since 40 can go into 6 multiple times, this fraction is going to be a mixed number.
40/6 in simplest form is 6 2/3. Hope I helped!
Answer:
c- translation
Step-by-step explanation:
it just moved a few points down
