1. You can set up 2 equations like so:
[tex]y = \frac{2}{3} \times x \\ x + y = 45[/tex]
You can then subsitute the y in the second equation with the first one.
[tex] \frac{2}{3}x + x = 45 \\ \frac{2}{3}x + \frac{3}{3} x = 45 \\ \frac{5}{3} x = 45[/tex]
And then you can multiply both sides by the reciprocal of 5/3.
[tex]x = 45 \times \frac{3}{5} \\ x = \frac{135}{5} \\ x = 27[/tex]
Then you can put this x into one of the equations to get the y like so:
[tex]y = \frac{2}{3} \times 27 \\ y = \frac{54}{3} \\ y = 18[/tex]
So your two numbers would be 18 and 27.
2. This is going to be similar to the first problem so I will cut down on the descriptions and just show the work.
[tex]y = 5 + x \\ y + x = 17[/tex]
Subsitute again.
[tex] x+ x + 5= 17 \\ 2x + 5 =17 \\ [/tex]
Subtract 5.
[tex]2x = 12 \\ x = 6[/tex]
Plug 6 back into one of the equations.
[tex]y = 6 + 5 \\ y = 11[/tex]
Your two numbers are 11 and 6.
3. This is a bit more complicated but the same logic applies here.
[tex]y = 8 + x \\ 3x + 2y = 26[/tex]
Subsitute again and solve for x.
[tex]3x + 2(x + 8) = 26 \\ 3x + 2x + 16 = 26 \\ 5x + 16 = 26 \\ 5x =10 \\ x = 2[/tex]
Then plug it in again.
[tex]y = 8 + 2 \\ y = 10[/tex]
Your two numbers are 2 and 10.
4. For this one, I'm going to have the $1 bills as y and the $5 bills as x.
[tex]y = 5x \\ x + y = 48[/tex]
Subsitute and solve for x.
[tex]x + 5x = 48 \\ 6x = 48 \\ x = 8[/tex]
Plug x back into an equation.
[tex]y = 5 \times 8 \\ y = 40[/tex]
Thus, Dan has 40 $1 bills and 8 $5 bills. I hope this helps.
