 My\:favourite paint\:comes\:in\:1L,\:2L,\:3L\:and\:5L\:tins.\:I\:bought\:13\:tins\:and\:a\:total\:of\:29L\:of\:paint.When\:I\:got\:home,\:I\:noticed\:that\:I\:had\:bought\:a\:different\:number\:of\:tins\:of\:each\:size\:but\:at\:least\:one\:of\:each.\:How\:many\:3L\:tins\:did\:I\:buy?

Let's define the variables for the number of tins you bought:

- \( x_1 \) = number of 1L tins
- \( x_2 \) = number of 2L tins
- \( x_3 \) = number of 3L tins
- \( x_4 \) = number of 5L tins

According to the problem, we have two key pieces of information:

1. The total number of tins is 13:
   \[
   x_1 + x_2 + x_3 + x_4 = 13
   \]

2. The total volume of paint is 29 liters:
   \[
   1x_1 + 2x_2 + 3x_3 + 5x_4 = 29
   \]

Additionally, we know that at least one tin of each size was bought, so:
\[
x_1 \geq 1, \quad x_2 \geq 1, \quad x_3 \geq 1, \quad x_4 \geq 1
\]

Now, let's solve this system of equations. Substituting the values to simplify:

Let \( y_1 = x_1 - 1 \), \( y_2 = x_2 - 1 \), \( y_3 = x_3 - 1 \), and \( y_4 = x_4 - 1 \). This substitution accounts for the fact that there is at least one tin of each size:

\[
(y_1 + 1) + (y_2 + 1) + (y_3 + 1) + (y_4 + 1) = 13
\]
Simplifies to:
\[
y_1 + y_2 + y_3 + y_4 = 9
\]

Similarly, for the volume equation:
\[
1(y_1 + 1) + 2(y_2 + 1) + 3(y_3 + 1) + 5(y_4 + 1) = 29
\]
Simplifies to:
\[
y_1 + 2y_2 + 3y_3 + 5y_4 = 19
\]

Now, we solve for \( y_1 \), \( y_

My favourite paint comes in 1L, 2L, 3L, and 5L tins. I bought 13 tins and a total of 29L of paint. When I got home, I noticed that I had bought a different number of tins of each size but at least one of each. How many 3L tins did I buy?

Learn how to solve a system of linear equations involving different sizes of paint tins. This problem involves finding the number of 3L tins purchased based on given totals of tins and paint volume. Suitable for students in Grades 7-9.

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