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Hi, can you please help me solve the simultanious equation with the steps of how to do it.

x^2 + y^2 = 25

3x + y = 5

Thanks.

Guest Jan 19, 2018

#3
+91773
+2

Hi, can you please help me solve the simultanious equation with the steps of how to do it.

$$x^2 + y^2 = 25\quad(1)\\ 3x + y = 5 \quad(2)$$

$$x^2 + y^2 = 25$$      This is a circle centre (0,0) with radius 5

$$y = -3x+5$$             and this line with gradient -3 and y intercept 5

So straight off I can see that one point of intersection is (0,5)

I know that there is one more.

I'm going square the second equ and then the first from it

$$(3x+y)^2=5^2\\ 9x^2+y^2+6xy=25\qquad (2a)\\ (2a)-(1)\\ 9x^2+y^2+6xy-x^2-y^2=25-25\qquad \\ 8x^2+6xy=0\qquad \\ 2x(4x+3y)=0\qquad \\ x=0 \qquad or \qquad 4x+3y=0\\ When\;\; x=0 \quad y=5 \quad \text{We know about that one}\\ 4x+3y=0\\ 4x=-3y\\ y=\frac{-4x}{3}\qquad (3) \\ \text{Now I can solve (2) and (3) simultaneously}\\ -3x+5=\frac{-4x}{3}\\ -9x+15=-4x\\ 15=5x\\ x=3\\ y=-3*3+5=-4\\ (3,-4)$$

So the points of intersection are (0,5) and  (3,-4)

And here I have checked my answer :)

Melody  Jan 19, 2018
Sort:

#3
+91773
+2

Hi, can you please help me solve the simultanious equation with the steps of how to do it.

$$x^2 + y^2 = 25\quad(1)\\ 3x + y = 5 \quad(2)$$

$$x^2 + y^2 = 25$$      This is a circle centre (0,0) with radius 5

$$y = -3x+5$$             and this line with gradient -3 and y intercept 5

So straight off I can see that one point of intersection is (0,5)

I know that there is one more.

I'm going square the second equ and then the first from it

$$(3x+y)^2=5^2\\ 9x^2+y^2+6xy=25\qquad (2a)\\ (2a)-(1)\\ 9x^2+y^2+6xy-x^2-y^2=25-25\qquad \\ 8x^2+6xy=0\qquad \\ 2x(4x+3y)=0\qquad \\ x=0 \qquad or \qquad 4x+3y=0\\ When\;\; x=0 \quad y=5 \quad \text{We know about that one}\\ 4x+3y=0\\ 4x=-3y\\ y=\frac{-4x}{3}\qquad (3) \\ \text{Now I can solve (2) and (3) simultaneously}\\ -3x+5=\frac{-4x}{3}\\ -9x+15=-4x\\ 15=5x\\ x=3\\ y=-3*3+5=-4\\ (3,-4)$$

So the points of intersection are (0,5) and  (3,-4)

And here I have checked my answer :)

Melody  Jan 19, 2018

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