Casey can buy 3 sandwiches and 5 cups of coffee for $26. eric can buy 4 sandwiches and 2 cups of coffee for $23. how much does one cup of coffee cost?

Let's identify what we are looking for in terms of variables.  Sandwiches are s and coffee is c.  Casey buys 3 sandwiches, which is represented then by 3s, and 5 cups of coffee, which is represented by 5c.   Those all put together on one bill comes to 26.  So Casey's equation for his purchases is 3s + 5c = 26.  Eric buys 4 sandwiches, 4s, and 2 cups of coffee, 2c, and his total purchase was 23.  Eric's equation for his purchases then is 4s + 2c = 23.  In order to solve for c, the cost of a cup of coffee, we need to multiply both of those bolded equations by some factor to eliminate the s's.  The coefficients on the s terms are 4 and 3.  4 and 3 both go into 12 evenly, so we will multiply the first bolded equation by 4 and the second one by -3 so the s terms cancel out.  4[3s + 5c = 26] means that 12s + 20c = 104.  Multiplying the second bolded equation by -3:  -3[4s + 2c = 23] means that -12s - 6c = -69.  The s terms cancel because 12s - 12s = 0s.  We are left with a system of equations that just contain one unknown now, which is c, what we are looking to solve for.  20c = 104 and -6c = -69.  Adding those together by the method of elimination (which is what we've been doing all this time), 14c = 35.  That means that c = 2.5 and a cup of coffee is $2.50.  There you go!

I'll say b) log(- that if not let me know what i can do