INFORMATION:
We know that:
- John sells plain cakes for $8 and decorated cakes for $12.
- On a particular day, John started with a total of 100 cakes, and sold all but 4.
- His sales that day totaled $800.
And we must find the number of plain cakes and decorated cakes that he sold that day.
STEP BY STEP EXPLANATION:
To find them, we can represent the situation using a system of equations
![\begin{cases}x+y={100-4...(1)} \\ 8x+12y={800...(2)}\end{cases}](https://img.qammunity.org/2023/formulas/mathematics/college/am3lvpju75w7txwnf6mlxdkxbg6hs51kjq.png)
Where, x represents the number of plain cakes that he sold and y represents the number of decorated cakes that he sold.
Now, we must solve the system:
1. We must multiply the equation (1) by -8
![\begin{gathered} -8(x+y)=-8(100-4) \\ -8x-8y=-768...(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b80ecxv53lwn3kspc3144jo6uirp68zaen.png)
2. We must add equations (2) and (3)
![\begin{gathered} 8x+12y=800 \\ -8x-8y=-768 \\ ---------- \\ 0x+4y=32 \\ \text{ Simplifying, } \\ 4y=32...(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/u8fl8xgw6wlfni4u4x5iw58d10ryb8w9yv.png)
3. We must solve equation (4) for y
![\begin{gathered} 4y=32 \\ y=(32)/(4) \\ y=8 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oo6nj7t4kqlspopv8yljqbo3bdgfimt1u4.png)
4. We must replace the value of y in equation (1) and then solve it for x
![\begin{gathered} x+8=100-4 \\ x=100-4-8 \\ x=88 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/byhcjybndr6nakp6z3niepksiyyvmyfy31.png)
So, we found that x = 88 and y = 8.
Finally, John sold 88 plain cakes and 8 decorated cakes.
ANSWER:
He sold 88 plain cakes and 8 decorated cakes that day.