Y=3x-9
0=3x-9
-3x=-9
x=3
X-int = 3
Answer:
The easiest way of solving this is to eliminate one of the variables. If you multiply each of the equations by a number so you can add the equations together and have only one variable left, you can solve the single variable equation. Try multiplying the first equation by 11, the second by -10 and adding them, thus eliminating the y variable. After this, substitute that value you have for x and solve for y. I hope this helps.
Step-by-step explanation:
Answer:
Step-by-step explanation:
8/15 + 2/5 = 8/15 + 6/15 + 14/15
-9y=-3x+216
y=1/3x-24
m=-3
y=-3x+16 is perpendicular to the line
Problem # 1
Not Factored: (5x^2 - 13x - 6)
Factored: (5x + 2 )(x - 3)
There is no real "work" to be shown for this, you can see that
1) Seperating 5x^2 into 5x and x will get you the equation in the form:
(5x ) (x ) = 5x^2 +
2) To complete this factor you need to guess which two numbers will add together to give you - 13x and multiply to form -6 (from the original unfactored equation). The numbers that will do this are 2 and 3
3) You can plug in negative or positives of those numbers to make sure they give you the exact results you need.
For example testing -2 and 3 you will get: (5x - 2) (x + 3 ) = 5(x^2) - 2x + 15x - 6 = 5x^2 +13x - 6. This is NOT the same as the unfactored equation. So you know -2 and 3 is the wrong choice. Choosing 2 and -3 will give you the answer instead.
As i said, there's no actual "work" to show this, you have to make guesses and try to factor it.
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Problem # 2
Not Factored: 4x^4 - 28x^3 + 48x^2
Factored: 4x^2 (x^2 - 7x + 12) =
4x^2 (x - 3) (x - 4)
To solve this, you factor out the 4x^2 from the entire equation. Then you can further factor the quadratic equation by seperating x^2 into (x )(x ) and guessing for which numbers will add to -7x and multiply to 12
