Step-by-step explanation
Let the number of each type of greeting cards be represented by x and y.
Since there are 6 greeting cards in total; it follows that:
![x+y=6----i](https://img.qammunity.org/2023/formulas/mathematics/college/g5s1k3p4dy1pw8convbcprx74w98ar23pv.png)
If x greeting card costs $2 each, then the total cost of that type of card = 2x
Also, if y greeting card costs $3 each, then the total cost of that type of card = 3y.
Therefore, if Lola paid $13 for the 6 greeting cards, it follows that:
![2x+3y=13----ii](https://img.qammunity.org/2023/formulas/mathematics/college/imck7vq2dg1gf0vz10nbgtjuvd3vfv6xdj.png)
Hence, the system of equations that represent this situation is:
![\begin{gathered} x+y=6-----i \\ 2x+3y=13---ii \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/g48tjjaql46fb7ki6zpzfzubsk89qkdgoq.png)
To solve the system of equations, use the elimination method.
Multiply (i) by 3
![\begin{gathered} x+y=6----i*3 \\ 3x+3y=18---iii \\ \text{Substact (}ii)\text{ from (}iii) \\ 3x-2x+3y-3y=18-13_{} \\ x=5 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/k1k7p4tqbtgzsvie89wzmgurkz4zksf53z.png)
To solve for y, put x = 5 into equation (i)
![\begin{gathered} \text{Recall (i)} \\ x+y=6----i \\ 5+y=6 \\ y=6-5 \\ y=1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ux8bv722nnlvmrdldseepk3ym50mjjapp9.png)
Hence, the solution to the system of equations is x = 5, and y = 1
The interpretation of the solution
x = 5 implies there are 5 greeting cards that cost $2 each.
y = 1 implies there is 1 greeting card that cost $3 each.