Math Help - equation system

1. equation system

A birdwatcher looking for robins and cardnials sees seven more cardinals than robins on saturday morning. The next day, she sees twice as many robins as she did the previous day, but only one-third of the number of cardnials. Her two-day total was 83. How many of each type did she see????

I got 60 robins
and 23 cardinals

thank you so much
armath

2. Hello, armath1

A birdwatcher looking for robins and cardinals sees seven more cardinals
than robins on Saturday morning.
The next day, she sees twice as many robins as she did the previous day,
but only one-third of the number of cardnials.
Her two-day total was 83.
How many of each type did she see?

I got 60 robins and 23 cardinals . . . . no
Organize the information in a chart . . .

On Saturday, she saw $R$ robins and $R+7$ cardinals.

$\begin{array}{c|c|c}
& \text{Sat.} & \text{Sun.} \\ \hline
\text{Cardinals} & R+7 & \\ \hline
\text{Robins} & R & \\ \hline
\end{array}$

On Sunday, she saw twice as many robins, $2R$,
and a third as many cardinals, $\frac{1}{3}(R+7)$
$\begin{array}{c|c|c}
& \text{Sat.} & \text{Sun.} \\ \hline
\text{Cardinals} & R+7 & \frac{1}{3}(R+7) \\ \hline
\text{Robins} & R & 2R \\ \hline
\end{array}$

She saw a total of 83 birds: . $(R+7) + \frac{1}{3}(R+7) + R + 2R \;=\;83$

Multiply by 3: . $3R + 21 + R + 7 + 3R + 6R \;=\;249$

. . . . $13R \;=\;221\quad\Rightarrow\quad R \:=\:17$

$\begin{array}{c|c|c|c}
& \text{Sat.} & \text{Sun.} & {\color{blue}\text{Total}} \\ \hline
\text{Cardinals} & 24 & 8 & {\color{blue}32}\\ \hline
\text{Robins} & 17 & 34 & {\color{blue}51}\\ \hline
\end{array}$

3. Originally Posted by armath
A birdwatcher looking for robins and cardnials sees seven more cardinals than robins on saturday morning. The next day, she sees twice as many robins as she did the previous day, but only one-third of the number of cardnials. Her two-day total was 83. How many of each type did she see????

I got 60 robins
and 23 cardinals

thank you so much
armath
I'm afraid that doesn't work.

Let r be the number of robins on the first day, and c the number of cardinals.

The we know that
$c = r + 7$

On the second day we have 2r robins and (1/3)c cardinals. We know that

$(r + c) + \left ( 2r + \frac{1}{3}c \right ) = 83$

So we have the two simultaneous equations:
$c = r + 7$
and
$(r + c) + \left ( 2r + \frac{1}{3}c \right ) = 83$

Try solving this and let's see what you get.

-Dan