How to solve this 2x+x-4y
2 answers:
Answer:
Step-by-step explanation:
If a <u>term</u> doesn't have a <u>coefficient</u> , it is considered that the coefficient is 1
Why is it considered that the coefficient is 1?
Remember that any term multiplied by 1 remains the same:
Step 1
The equality can be read in the other way as well , so any term can be written as a product of 1 and itself:
Step 2
Usually, we don't need to write multiplication sign between the coefficient and the variable , so the simpler form is:
That is why we can write the term without the coefficient as a term with coefficient 1.
Let's go back to your problem now
Collect l<u>ike terms </u>by adding their <u>coefficients</u>
Add the numbers
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If you are asking what is the graph of y = 3x^2 -2x+1.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.
Then, the attached file would be the answer.
To check, b^2 - 4(a)(c), for each equation and use these facts:
If b^2 - 4(a)(c) = 0, there is only one real root meaning, the graph touches the x-axis only in one point.
If b^2 - 4ac > 0, there are two real roots meaning, the graph touches the x-axis in two different points.
If b2 - 4ac < 0, there are no real roots then the graph does not touch the x-axis. This would be the case for y = 3x^2 - 2x + 1.
Solution:
(-2)^2 -4(3)(1) = 4 - 12 = -8 < 0 will result in not real roots.