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2 years ago

7x-3y=-12. Name 2 possible answers for (x,y)

Mathematics

1 answer:

anzhelika [568]2 years ago

6 0

Answer:

x = (-12/7 , 0)

y= (0,4)

Step-by-step explanation:

Plug y=0 into the equation and solve the resulting equation 7x=−12 for x

7x = -12

/7 /7

x = -12/7 and y =0

Plug x=0 into the equation and solve the resulting equation −3y=−12 for y

-3y = -12

/-3 -3

Y = 4 and X = 0

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Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0