# Thread: Solve Simultaneous Equations Graphically?

1. ## Graphical solutions of simultaneous equations

Hello Sailee316
Originally Posted by Sailee316
Hi. I have this question on my assignment:

2. Solve graphically:

(a) 2x – 3y = 11/12 and x + y = -7/12

(b) y = 30 – 6x + 2x2 and y = 2(x + 11)

Does this mean I have to do 4 seperate graphs or just 2? I did some research and found I need to reaarange and find points to plot but im having trouble rearrainging the four different equations. Thanks
These equations are much easier to solve algebraically than graphically, but if that's what the question says...

(a)

$\displaystyle \left\{\begin {array}{l}2x - 3y =\frac{11}{12}\\\,\\x+y = -\frac{7}{12}\end{array} \right.\$

Re-arrange:

$\displaystyle \left\{\begin{array}{l} y = \frac{1}{3}\left(2x - \frac{11}{12}\right)\\\,\\y=-x-\frac{7}{12}\end{array} \right.\$

Then re-write using approximate decimal values, so that you can use a calculator:

$\displaystyle \left\{\begin{array}{l} y = 0.333\left(2x - 0.917\right)\\\,\\y=-x-0.583\end{array} \right.\$

Trial and error will show you which values of $\displaystyle x$ to use in your two formulae, to work out values of $\displaystyle y$. But if you use values of $\displaystyle x$ from $\displaystyle -1$ to $\displaystyle +1$ in steps of $\displaystyle 0.2$, that should work OK.

Plot the two graphs on the same diagram (they're both straight lines), and read off their point of intersection. (The answer, that I found algebraically, is approx. $\displaystyle x = -0.167, y = -0.417$.)

(b) I assume that the first equation is $\displaystyle y = 30-6x+2x^2$. Again, trial and error will show you which values of $\displaystyle x$ to use. If you try $\displaystyle x=0$ to $\displaystyle x=4$ in steps of $\displaystyle 0.5$, that should be OK.

Again, plot the two graphs on a single diagram. This time, one is a curve and the other turns out to be a tangent to the curve. Read off the values of $\displaystyle x$ and $\displaystyle y$ where the tangent touches the curve. An algebraic method tells me it is $\displaystyle x = 2, y = 26$.

Grandad

2. Ok I have used a calculator for the second but this calculator:

Graphing Calculator
Does not seem to like these equations:

2x – 3y = 11/12

x + y = -7/12

Could you tell me a way to enter both these equations in a way this calculator understands? Rearrange possible? Thanks

3. Originally Posted by Sailee316
Ok I have used a calculator for the second but this calculator:

Graphing Calculator
Does not seem to like these equations:

2x – 3y = 11/12

x + y = -7/12

Could you tell me a way to enter both these equations in a way this calculator understands? Rearrange possible? Thanks
To get rid of the fraction on the right hand side you could multiply both sides by 12:

$\displaystyle 2x -3y = \frac{11}{12}$ becomes $\displaystyle 24x -36y = 11$

Then do the same for the second one.

Hope this helps

4. So how do I do this for the second one. Thanks

5. You would do the same as I have done above. I'm sure that you can times x and y by 12?

6. So for (a) when solving algebraically you get

x = -0.167, y = -0.417

obviusly you cant possibly find that intersection point to that degree of acuraccy when reading off a graph so what should I write as my answer?? And still get the question correct? Thanks for you help that was greatly appreciated.

Also which from where to where would you draw your graph im thinking -30 to 30 but wont the lines just look like 0 if you do this?

If I done -4 to 4 I think I will be able to read off an accurate intersection but the only reason I know that is that I have been using online calculators oops. So what im asking is
what kind of graph would you draw to find an intersection for:

24x-36y=11
12x+12y=-7

[/COLOR]
Originally Posted by Grandad
Hello Sailee316These equations are much easier to solve algebraically than graphically, but if that's what the question says...

(a)

$\displaystyle \left\{\begin {array}{l}2x - 3y =\frac{11}{12}\\\,\\x+y = -\frac{7}{12}\end{array} \right.\$

Re-arrange:

$\displaystyle \left\{\begin{array}{l} y = \frac{1}{3}\left(2x - \frac{11}{12}\right)\\\,\\y=-x-\frac{7}{12}\end{array} \right.\$

Then re-write using approximate decimal values, so that you can use a calculator:

$\displaystyle \left\{\begin{array}{l} y = 0.333\left(2x - 0.917\right)\\\,\\y=-x-0.583\end{array} \right.\$

Trial and error will show you which values of $\displaystyle x$ to use in your two formulae, to work out values of $\displaystyle y$. But if you use values of $\displaystyle x$ from $\displaystyle -1$ to $\displaystyle +1$ in steps of $\displaystyle 0.2$, that should work OK.

Plot the two graphs on the same diagram (they're both straight lines), and read off their point of intersection. (The answer, that I found algebraically, is approx. $\displaystyle x = -0.167, y = -0.417$.)

(b) I assume that the first equation is $\displaystyle y = 30-6x+2x^2$. Again, trial and error will show you which values of $\displaystyle x$ to use. If you try $\displaystyle x=0$ to $\displaystyle x=4$ in steps of $\displaystyle 0.5$, that should be OK.

Again, plot the two graphs on a single diagram. This time, one is a curve and the other turns out to be a tangent to the curve. Read off the values of $\displaystyle x$ and $\displaystyle y$ where the tangent touches the curve. An algebraic method tells me it is $\displaystyle x = 2, y = 26$.

Grandad

7. Originally Posted by Sailee316
Ok I have used a calculator for the second but this calculator:

Graphing Calculator
Does not seem to like these equations:

2x – 3y = 11/12

x + y = -7/12

Could you tell me a way to enter both these equations in a way this calculator understands? Rearrange possible? Thanks
You've already been given the re-arrangement - In post #16 Grandad made y the subject for each equation.

Page 2 of 2 First 12