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