# Thread: Solving Simultaneous Equations By Plotting A Graph

1. ## Solving Simultaneous Equations By Plotting A Graph

Can anybody help with the following problem, i'm not very knowledgeable on this subject.

Solve the following simultaneous equations by plotting the graphs between x = -4 and x = +3

y = 3x² + 2x - 5 and y = 2x +10

Use your graph to solve 3x² + 2x - 5 = 0

??????????

2. Originally Posted by c00ky
Can anybody help with the following problem, i'm not very knowledgeable on this subject.

Solve the following simultaneous equations by plotting the graphs between x = -4 and x = +3

y = 3x² + 2x - 5 and y = 2x +10

Use your graph to solve 3x² + 2x - 5 = 0

??????????
I have attached the graph of both equations for values between x=-4 and x=3 but it's impossible to find exact points of intersection by looking at the graph for y = 3x² + 2x - 5 (blue curve) and y = 2x +10 (red line).
I suggest you do it by solving system by substitution method.

Also finding when exact 3x² + 2x - 5 equals to 0 by looking at graph is not possible. It can be seen from graph that one solution is x=1 but other solution is somewhere between -1,5 and -2. So that too equation must be solved.

Graph for both tasks is useless here.

3. Can anybody expand on this answer. I'm unsure of how i'm supposed to solve it.

Thanks

4. Originally Posted by c00ky
Can anybody expand on this answer. I'm unsure of how i'm supposed to solve it.

Thanks
1)
$\displaystyle \begin{array}{l} y = 3x^2 + 2x - 5 \\ y = 2x + 10 \\ \end{array}$

You equal both y's and solve equation for x:
$\displaystyle \begin{array}{l} 3x^2 + 2x - 5 = 2x + 10 \\ 3x^2 - 15 = 0 \\ 3x^2 = 15 \\ x^2 = 5 \\ x = \pm \sqrt 5 \\ \end{array}$

Now we find y for both values of x:
$\displaystyle \begin{array}{l} y = 2\sqrt 5 + 10 \\ \\ y = - 2\sqrt 5 + 10 \\ \end{array}$

So solutons are (or points of intersection):
$\displaystyle (\sqrt 5 ,2\sqrt 5 + 10) \wedge ( - \sqrt 5 , - 2\sqrt 5 + 10)$

2) $\displaystyle 3x^2 + 2x - 5 = 0$

This is quadratic equation and it can be solved by applying quadratic formula.
Although this expression can be factored so I will not use quadratic formula:
$\displaystyle (3x + 5)(x - 1) = 0$

So solutions are:
$\displaystyle \begin{array}{l} 3x + 5 = 0 \wedge x - 1 = 0 \\ x = - \frac{5}{3} \wedge x = 1 \\ \end{array}$