Question

solve the simultaneous equations;x+3y+9=o and y=x^2-3x-4

Answer 1

Step-by-step explanation:

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Answer 2

Given:

Two equations-

• x+3y+9= 0----------(1)
• y= x²-3x-4----------(2)

To Find:

Intersection point of given curves or values of x and y for which both curves intersect at a point

Solution:

Equation (1) can also be written as

$x+3y=-9$

$3y=-x-9$

$y=\dfrac{-x-9}{3}-------(3)$

On equating (2) and (3), we get

$\longrightarrow\mathrm{x^{2}-3x-4=\dfrac{-x-9}{3}}$

$\longrightarrow\mathrm{3(x^{2}-3x-4)=-x-9}$

$\longrightarrow\mathrm{3x^{2}-9x-12=-x-9}$

$\longrightarrow\mathrm{3x^{2}-9x-12+x+9=0}$

$\longrightarrow\mathrm{3x^{2}-8x-3=0}$

$\longrightarrow\mathrm{3x^{2}-9x+x-3=0}$

$\longrightarrow\mathrm{3x(x-3)+1(x-3)=0}$

$\longrightarrow\mathrm{(3x+1)(x-3)=0}$

$\mathrm{x=\dfrac{-1}{3},3}$

On putting values of x in (3), we get

Case-1: When $x=\dfrac{-1}{3}$ 

$\longrightarrow\mathrm{y=\dfrac{-\bigg(\dfrac{-1}{3}\bigg)-9}{3}}$

$\longrightarrow\mathrm{y=\dfrac{\dfrac{1}{3}-9}{3}}$

$\longrightarrow\mathrm{y=\dfrac{\dfrac{1-27}{3}}{3}}$

$\longrightarrow\mathrm{y=\dfrac{-26}{9}}$

Case-2: When x= 3

$\longrightarrow{\mathrm{y=\dfrac{-3-9}{3}}}$

$\longrightarrow\mathrm{y=\dfrac{-12}{3}}$

$\longrightarrow\mathrm{y=-4}$