Let flower bouquet be [tex]f[/tex] and vase be [tex]v[/tex]
Form two equations as follow
[tex]3f+5v=25.75[/tex] ---> Equation 1
[tex]8f+2v=29[/tex] ----> Equation 2
Using the elimination method:
Choose which variable, either [tex]f[/tex] or [tex]v[/tex] that we want to eliminate first. Say we choose [tex]v[/tex]
The next step is we need to make the constants of [tex]v[/tex] to be the same value. At the moment we have [tex]5v[/tex] and [tex]2v[/tex]. Using the concept of common multiple, we will multiply [tex]5v[/tex] by [tex]2[/tex] and the [tex]2v[/tex] by 5 to get both [tex]v[/tex]'s with a constant 10
The rule of the Algebraic equation is: if we multiply one term by [tex]k[/tex], then we need to do the same with all the other terms in the equations. So now we have
([tex]3f+5v=25.75[/tex])×2 ⇒[tex]6f+10v=51.5[/tex]
([tex]8f+2v=29[/tex])×5⇒[tex]40f+10v=145[/tex]
Now we can eliminate [tex]v[/tex] by subtracting Equation 1 from Equation 2 (it would also work the other way around)
Equation 2 - Equation 1
[tex]40f-6f+10v-10v=145-51.5[/tex]
[tex]34f=93.5[/tex]
[tex]f=2.75[/tex]
Now we've worked out the price of one flower bouquet, we can substitute back [tex]f=2.75[/tex] into either equation 1 or equation 2 to work out the price of one vase
[tex]3(2.75)+5v=25.75[/tex]
[tex]5v=25.75-8.25[/tex]
[tex]5v=17.5[/tex]
[tex]v=3.5[/tex]
Hence, the price of one bouquet is $2.75 and one vase is $3.50
