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3 years ago

Pls i need it like now

Mathematics

1 answer:

schepotkina [342]3 years ago

7 0

You need to put the x or y value in the other equation ;
For example excersize a
It given that
Y=x^2+3x-1
Then you put it on the first equation :
x+x^2+3x-1=4
And then you solve it
x+x^2+3x-1-4=0
4x+x^2-5=0
We will make the order to seem easier
x^2+4x-5=0
x1=-5
x2=1
Then you put the x1,2 that you found on the second equation :
y=(-5)^2+3*(-5)-1=9
y=1^2+3*1-1=3
For summary :
x1=-5
x2=1
y1=9
y2=3

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$x\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

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Substituting the values, we have it that:

$\begin{gathered} x\text{ = }\frac{-149\pm\sqrt[]{149^2-4(-16)(108)}}{2(-16)} \\ \\ x\text{ = }\frac{-149\pm\sqrt[]{29113}}{-32}\text{ } \\ \\ x\text{ = }\frac{-149-170.625\text{ }}{-32} \\ or\text{ } \\ x\text{ = }\frac{-149_{}+170.625}{-32} \end{gathered}$

Now, we proceed to get the individual x-values:

$\begin{gathered} x\text{ = }\frac{-149-170.625}{-32}\text{ = 9.99 } \\ or \\ x\text{ = }\frac{-149+170.625}{-32}\text{ = -0.68} \end{gathered}$

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