# Math Help - exam help

1. ## exam help

um could anyone plz help me with this queston thats on an exam tommorow....and i need at least a B+....

y= 4√(x+4) -2
y= ⅔x +7

simultaneous equation....i need to sketch a graph, label intercepts and state their value

2. have you tried to graph that on a calculator or something.

The second one is a straight line

Originally Posted by Awsom Guy
The second one is a straight line

the intercepts for the second equation is y is 7 and x is 10. I just graphed this into my calculator so it is right and the other equation is crazy I am still trying to work it out.

3. If I say this could you do it:
4sqrt(x+4) - 2 (1)
2/3x + 7 (2)

Make equation 1 like this solve it first:
4(x^1/2)+2) -2
4x^1/2 + 8 -2
4(x^1/2) + 6

Therefore we get our 2 equations:
4(x^1/2) + 6
(2/3)x + 7
Can you solve from there...

4. Originally Posted by Awsom Guy
If I say this could you do it:
4sqrt(x+4) - 2 (1)
2/3x + 7 (2)

Make equation 1 like this solve it first:
4(x^1/2)+2) -2
4x^1/2 + 8 -2
4(x^1/2) + 6

Therefore we get our 2 equations:
4(x^1/2) + 6
(2/3)x + 7
Can you solve from there...

ummm is "^" square root sign?

5. yes like 2^2=4

6. 1) Sketch a graph : just plug in $x = -3, -2, -1, 0, 1, 2, 3$ and get the corresponding $y$ values. Then ... plot. Note that for the second one only two points are required since it is the equation of a line.

2) Label intercepts : that shouldn't be too hard, for the x-intercepts just set $y = 0$ and solve for $x$ in both cases, for the y-intercepts just set $x = 0$ and solve for $y$ in both cases. You therefore get their values which you can then sketch.

As you mentioned "simultaneous equations" I assume you need to solve the system at some point (know where the two curves intersect). You can apply the following method :

Just set $4 \sqrt{x + 4} - 2 = \frac{2}{3} x + 7$ and solve for $x$ :

$4 \sqrt{x + 4} - 2 = \frac{2}{3} x + 7$

$4 \sqrt{x + 4} = \frac{2}{3} x + 9$

$\left ( 4 \sqrt{x + 4} \right ) ^2 = \left ( \frac{2}{3} x + 9 \right ) ^2$

$16(x + 4) = \left ( \frac{2}{3} x \right )^2 + \frac{36}{3} x + 9^2$

$16x + 64 = \frac{4}{9} x^2 + 12x + 81$

$0 = \frac{4}{9} x^2 - 4x + 17$

$9 \times 0 = 9 \left ( \frac{4}{9} x^2 - 4x + 17 \right )$

$0 = 4 x^2 - 36x + 153$

$4 x^2 - 36x + 153 = 0$

Calculate the discriminant :

$\Delta = b^2 - 4ac = (-36)^2 + 4 \times 4 \times 153 = -1152$

It is negative therefore this equation has no real solution for $x$, thus the two curves (the curve and the line) never meet (never intersect).

Does it make sense ?

7. Originally Posted by Awsom Guy
yes like 2^2=4
thank u so much yet again then!