let chairs be [tex]c[/tex] and table be [tex]t[/tex]
We form two equations from the information given so we can solve simultaneously
Eq 1 ⇒ [tex]5c+3t=35[/tex]
Eq 2 ⇒ [tex]7c+9t=91[/tex]
We can either use the method of elimination or substitution. We will use the elimination method for this one
Let us eliminate the term [tex]c[/tex]. We need to make the two constants the same. We have [tex]5c[/tex] and [tex]7c[/tex] and we can make them both as [tex]35c[/tex].
We will multiply each term in Eq 1 by 7 to obtain [tex]35c+21t=245[/tex]
We will multiply each term in Eq 2 by 5 to obtain [tex]35c+45t=455[/tex]
Now we subtract Eq 2 from Eq 1 to obtain
[tex](35c-35c)+(21t-45t)=(245-455)[/tex]
[tex]-24t=-210[/tex]
[tex]t= \frac{210}{24}=8.75 [/tex]
So the price of one table is $8.75
Substitute 8.75 into either Eq 1 or Eq 2 to obtain the price for one chair. Let's use Eq 1
[tex]5c+3(8.75)=35[/tex]
[tex]5c+26.25=35[/tex]
[tex]5c=35-26.25[/tex]
[tex]5c=8.75[/tex]
[tex]c=1.75[/tex]
So the price of one chair is $1.75
