y = 5 y = 3x - 4 PLEASE HELP ME WITH 3 QUESTIONS THAT'S ALL JUST 3!!

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y = 5 y = 3x - 4 PLEASE HELP ME WITH 3 QUESTIONS THAT'S ALL JUST 3!!

Mathematics
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here we can write this: \[\Large 3x - 4 = 5\]
Okay
now I add 4 to both sides: \[\Large 3x - 4 + 4 = 5 + 4\] please continue

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Other answers:

3,9
is 3,9 right or is it wrong what is it , I'm really frustrated so ....
after a simplification, we can write: \[\Large 3x = 9\]
now, please divide by 3, both sides
what sides am I dividing it by
I got 3
hint: |dw:1437154156003:dw|
that's right!
so is it 3,3
it is x=3
I have to write it as an ordered pair
ok! we have: x=3 and y=5, so we can write: (3,5)
Find the solution(s) to the system, if one or more exist. Write your answer as an ordered pair. Round to the nearest hundredth, if necessary. \begin{array}{l}y = {x^2} + 4x - 3\\x = 1\end{array} this is the other one, I have one more I need help on
we have to substitute x=1 into the first equation, so we can write: \[y = {1^2} + 4 \times 1 - 3 = ...?\]
okay
please continue, what is y?
42
that;s what I got for y
hint: \[y = {1^2} + 4 \times 1 - 3 = 1 + 4 - 3 = ...?\]
I got 3, 2
we have: y=2 and x=1 so we can write: (1,2)
is that the answer?
yes!
Explain how to find the solution(s) to the system. Give the solutions(s) as ordered pairs rounded to the nearest hundredth, if necessary. \begin{array}{l}y = - 0.5{x^2} + x + 6\\y = 3.25\end{array} this is the last question I need help on I need help finding the ordered pairs
we have to apply the transitive property, so we can write: \[\Large - 0.5{x^2} + x + 6 = 3.25\]
ok
it is a quadratic equation, since we can rewrite it as follows: \[\Large 0.5{x^2} - x - 2.75 = 0\]
ok
or using fraction instead decimal numbers, we get: \[\Large \frac{{{x^2}}}{2} - x - \frac{{11}}{4} = 0\]
ok
now, I multiply both sides by 4, so I can write: \[\Large \begin{gathered} 4 \times \left( {\frac{{{x^2}}}{2} - x - \frac{{11}}{4}} \right) = 0 \times 4 \hfill \\ 4 \times \frac{{{x^2}}}{2} + 4 \times \left( { - x} \right) + 4 \times \left( { - \frac{{11}}{4}} \right) = 0 \hfill \\ 2{x^2} - 4x - 11 = 0 \hfill \\ \end{gathered} \]
ok
note that equation is an equation like this: \[\Large a{x^2} + bx + c = 0\] where: \[\Large \begin{gathered} a = 2 \hfill \\ b = - 4 \hfill \\ c = - 11 \hfill \\ \end{gathered} \]
okay
so we can apply the standard formula: \[\Large x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}}\]
in order to find the values of x
okay
after a simple substitution, we get: \[\large \begin{gathered} x = \frac{{ - b \pm \sqrt {{b^2} - 4 \times a \times c} }}{{2 \times a}} = \frac{{4 \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \times 2 \times \left( { - 11} \right)} }}{{2 \times 2}} = \hfill \\ \hfill \\ = \frac{{4 \pm \sqrt {16 + 88} }}{4} \hfill \\ \end{gathered} \] please continue
I got 7
hint: we have 2 values for x: \[\Large \begin{gathered} {x_1} = \frac{{4 + \sqrt {16 + 88} }}{4} = ...? \hfill \\ \hfill \\ {x_2} = \frac{{4 - \sqrt {16 + 88} }}{4} = ...? \hfill \\ \end{gathered} \]
42
hint: \[\begin{gathered} {x_1} = \frac{{4 + \sqrt {16 + 88} }}{4} = \frac{{4 + \sqrt {104} }}{4} = \frac{{4 + 10.2}}{4} = \frac{{14.2}}{4} = ...? \hfill \\ \hfill \\ {x_2} = \frac{{4 - \sqrt {16 + 88} }}{4} = \frac{{4 - \sqrt {104} }}{4} = \frac{{4 - 10.2}}{4} = \frac{{ - 6.2}}{4} = ...? \hfill \\ \end{gathered} \]
14.24 -6.24
you have to divide by 4, namely 14.20/4=...? -6.20/4=...?
3.55 , 1.55
-1.55
we have: x1= 3.55 x2=-1.55
ok!
ok
so the requested pairs, are: (3.55, 3.25) (-1.55, 3.25)
Thank you
i need two that I wanted you to check
ok!
Find the solution(s) to the system, if one or more exist. Write your answer as an ordered pair. \begin{array}{l}y = 3{x^2} + 2x + 4\\x = 4\end{array} I have 156,4 as my answer
substituting x=4 into the first equatiuon, we get: \[y = 3 \times {4^2} + 2 \times 4 + 4 = 3 \times 16 + 8 + 4 = 48 + 12 = 60\]
so the requested pair is: (4, 60)
one of the answer they have is (4,60) and (60 ,4)
thank you
:)
the one with 3,5 wasn't right
right
please what question do you refer to?

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