Answer:
[tex]\boxed {\boxed {\sf Haircut: \$47 \ and \ Color: \$ 100}}[/tex]
Step-by-step explanation:
Let's set up a system of equations. (1 equation for yesterday and 1 for today).
- Let h = haircut and c= color
[tex]2h+1c= 194[/tex] (Yesterday: 2 haircuts and 1 color for $194).
[tex]2h+4c= 494[/tex] (Today: 2 haircuts and 4 colors for $494).
Notice that both equations have a 2h. If we subtract the two equations, the 2h will cancel and leave us with one variable, c.
[tex]\ \ (2h+1c=194) \\- (2h+4c=494)[/tex]
[tex]\ \ 1c=194 \\- (4c=494)[/tex]
[tex]-3c= -300[/tex]
Since we are solving for c, we must isolate the variable. It is being multiplied by -3. The inverse of multiplication is division. Divide both sides by -3.
[tex]-3c/-3=-300/-3 \\c= -300/-3 \\c=100[/tex]
Now we have to solve for h. Plug 100 in for c in the first equation.
[tex]2h+1(100)= 194 \\2h+100=194[/tex]
100 is being added, so we must subtract 100 from both sides.
[tex]2h+100=194-100 \\2h=94[/tex]
h is being multiplied by 2, so we divide both sides by 2.
[tex]2h/2=94 \\h=47[/tex]
A haircut costs $47 and a color costs $100.
