How long an investment will take to compund to a target amount 
A man with $\$20,000$ to invest decides to diversify his investments by placing $\$10,000$ in an account that earns $7.2\%$ compounded continuously and $\$10,000$ in an account that earns $8.4\%$ compounded annually. Use graphical approximation methods to determine how long it will take for his total investment in the two accounts to grow to $\$35,000$.

Teacher said the answer is: $7.3$ years
But i don't know how to solve it.
Please help me. 
FORMULAS:
Compounded Continuously: $A=Pe^{rt}$
Compounded Annually: $A=P(1+r)^t$
$A$ - Money after t years 
$P$ - Principal amount invested
$r$ - rate of investment 
$t$ - time in years 
$e$ - the value $e$ used in $\ln$
 A: Everything you need is in the graphic per the instructions.
As mentioned in the comments and shown in the graphic, you could also solve for $t$ using numerical methods like Newton's Method.
The equation becomes:
$$10000e^{.072 t}+10000(1+0.084)^t = 35000 \implies e^{.072 t}+(1.084)^t = 3.5$$
The result is $t = 7.32505 ~~\mbox{years}$ (see update below for more details).
Here is the graphical representation (look at the point for the vertical line):

Update
We can solve this numerically (Fixed Point Iteration, Secant Method, Newton's Method ... Others). Using Newton's Method, we have the iteration formula:
$$t_{n+1} = t_n- \dfrac{f(t)}{f'(t)} = t_n - \dfrac{e^{0.072 t_n} + 1.084^{t_n} -3.5}{0.0806579 \times 1.084^{t_n} + 0.072e^{0.072 t_n}}, ~ t_0 = -22$$
This converges in $13$ steps to $t = 7.325046477477088$.
A: The \$10,000 account at 7.2% interest gets you $10000e^{0.072t}$ dollars in $t$ years. The other \$10,000 investment gets you $10000(1+0.084)^t$ dollars in $t$ years. 
So in $t$ years, you have $10000e^{0.072t}+10000(1+0.084)^t$ total dollars. We can make this a bit easier by factoring out the $10000$, so the amount of money in $t$ years, $M(t)$, we have is $M(t)=10000(e^{0.072t}+1.084^t)$. 
We want to know in how many years will this be \$35,000. That is, we want $M(t)=35,000$. So
$$
10000(e^{0.072t}+1.084^t)=35,000
$$
Divide by the 10,000 to get
$$
e^{0.072t}+1.084^t=3.5
$$
Since your teacher wants this graphically, go into the calculator (I'll assume you have a TI-83 or TI-84) and graph the function $e^{0.072t}+1.084^t$ as Y1. Then for Y2 3.5. Then when you graph them, look where they intersect. That $x$ value where the lines touch will be the amount of years you want. Be careful about choosing an appropriate window! (But since we divided by the 10,000, this shouldn't be a problem - the only reason I did that). You can do this without having to 'guess' by using the 2nd - CALCULATE - then INTERSECT to find the intersection of the curves and make the calculator do all the work.
To get an idea of what you should see, trying looking at this plot here.
A: Using pencil, paper, eraser and a hand calculator that has exponential functions calculate and plot the graph, taking $1,000 as convenient unit :
$$ A(n) = 10 ( 1 + .072)^n + 10 ( 1 + .084)^n $$
EDIT1:
corrected (oversight error) to include continuous compounding:
$$ A(n) = 10\  e^{0.072 \, t} + 10\ ( 1 + .084)^n $$
for n upto 8 years. It takes ~7.32505 years to reach 35.
