# Samantha invests $1000 at 6% per annum compounded quarterly. Mark invests $1200 at 5.5% per annum compounded quarterly. When will the balances in their accounts be equal?

Aug 31, 2016

The balances in the accounts will be equal in $37$ years.

#### Explanation:

For problems such as these, we use the formula $A = P {\left(1 + \frac{r}{n}\right)}^{n t}$, where P is the amount of money you begin with, $r$ is the rate of interest, $n$ is the frequency the interest is compounded and A is the final amount. $t$ is the time in years.

We have to write a system of equations and solve for $A$ and for $t$.

Equation 1:

$A = 1000 {\left(1 + \frac{0.06}{4}\right)}^{4 t}$

Equation 2:

$A = 1200 {\left(1 + \frac{0.055}{4}\right)}^{4 t}$

Substitute:

$1200 {\left(1 + \frac{0.055}{4}\right)}^{4 t} = 1000 {\left(1 + \frac{0.06}{4}\right)}^{4 t}$

$1200 {\left(1.01375\right)}^{4 t} = 1000 {\left(1.015\right)}^{4 t}$

$1200 {\left(1.01375\right)}^{4 t} - 1000 {\left(1.015\right)}^{4 t} = 0$

Solve using a graphing calculator. If you're using a TI-83 or a TI-84, change the t's to x's. Enter ${y}_{1} = 1200 {\left(1.01375\right)}^{4 x} - 1000 {\left(1.015\right)}^{4 x}$ and ${y}_{2} = 0$. Then press CALC$\to$ Intersect.

You will of course want a positive intersection. Once you have moved your cursor sufficiently, press ENTER twice before the calculator will say guess? followed by INTERSECTION.

The result it gives you should say $36.99$, or $37$ years.

Hopefully this helps!