Let
x = number of school bags
y = number of sweaters
z = number of color pencils.
They want to buy 200 items translates to
[tex]x+y+z=200[/tex]With a total $2800, a price of $20 on school bags, $25 on sweaters and $5 on color pencils translates to
[tex]20x+25y+5z=2800[/tex]They must order as many color pencils as school bags and sweaters combined, translates to
[tex]z=x+y[/tex]Therefore, we have the following system
[tex]\begin{gathered} x+y+z=200\text{ first equation} \\ 20x+25y+5z=2800\text{ second equation} \\ z=x+y\text{ third equation} \end{gathered}[/tex]Substitute the z of the third equation to the first and second equation
[tex]\begin{gathered} x+y+z=200\text{ (first equation)} \\ x+y+(x+y)=200 \\ 2x+2y=200\text{ (fourth equation)} \\ \\ 20x+25y+5(x+y)=2800 \\ 20x+25y+5x+5y=2800 \\ 25x+30y=2800\text{ (fifth equation)} \end{gathered}[/tex]Solve the system of fourth and fifth equation using elimination method.
[tex]\begin{gathered} \text{Multiply the fourth equation by 15} \\ 15(2x+2y=200)\Longrightarrow30x+30y=3000 \\ \\ \text{Then subtract it by the fifth equation} \\ 30x+30y=3000 \\ -(25x+30y=2800) \\ ------------- \\ 5x=200 \\ \\ \frac{5x}{5}=\frac{200}{5} \\ x=40 \end{gathered}[/tex]Substitute the value of x to the fourth equation and solve for y
[tex]\begin{gathered} 2x+2y=200 \\ 2(40)+2y=200 \\ 80+2y=200 \\ 2y=200-80 \\ 2y=120 \\ \frac{2y}{2}=\frac{120}{2} \\ y=60 \end{gathered}[/tex]Finally, substitute the value of x and y to the third equation to solve for z
[tex]\begin{gathered} z=x+y \\ z=40+60 \\ z=100 \end{gathered}[/tex]Since y is the number of sweater, the number of sweaters that they have ordered is 60.