From the work above, I see that you have a strong understanding with two variable equations. You can identify two variables and write two equations that involve the variables. I noticed that you might need some guidance when solving these equation. Through these example I hope I am helpful in my explanations.
1. You did a good job with the variables and the equations. All you need to do is solve. There are two ways to solve systems of equations. You can solve through substitution or though elimination. I will give you the solution through the "elimination strategy" in this problem.
5s+2z=1.32
We want to get the "z" in this second equation to "2z".
3s+z=0.75 multiplied by a factor of 2 becomes 6s+2z=1.5 (The equation is still true because we applied the multiplication to all the terms.)
Now, let subtract the equations:
6s+2z=1.5
-(5s+2z=1.32)
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s=0.18 (notice that the z variables cancel out)
Lastly, plug in the value we got for s to find z in any equation.
3s+z=0.75
3(0.18)+z=0.75
z=0.21
answer: squash: $0.18
zucchini:$0.21
2. You did a good job finding variable for the unknowns. Let's make a few equations with the information provided.
8x+10y=80
9x+5y=65
Now, let us actually solve by elimination again because we can change 5y into 10y easily.
9x+5y=65 becomes 18x+10y=130 (multiplied by 2)
Now we can eliminate one of the variables through this strategy.
18x+10y=130
-(8x+10y=80)
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10x=50
x=5
Again, let us find y.
9x+5y=65
9(5)+5y=65
5y=20
y=4
answer: Jude - $5 an hour
Ben $4 an hour
3. Alright a money problem and time to showcase the other method: substitution. But first, we need to create and identify the variables.
d-number of dimes
q-number of quarters
Next, we create two equations like always based on the situation.
d+q=40
0.1d+0.25q=7.6 (dimes=$0.10 and quarters=$0.25)
Okay, time for elimination.
d+q=40 ----> d=40-q (just dividing by q on both sides)
Now for the aha moment.
d=40-q
0.1d+0.25q=7.6
We input "40-q" in the place of d in the second equation because 40-q equals d.
0.1(40-q)+0.25q=7.6 (One variable now? Now I can solve this easily!)
4-0.1q+0.25q=7.6
0.15q=3.6
q=24
Find d now.
d+q=40
d+24=40
d=16
answer: 24 quarters and 16 dimes
4. Almost the same problem, I will solve it more quickly this time using substitution.
d-number of dimes
q-number of quarters
d+q=24 ----> d=24-q
0.1d+0.25q=3.6
0.1(24-q)+0.25q=3.6
2.4-0.1q+0.25q=3.6
0.15q=1.2
q=8
d+q=24
d+8=24
d=16
answer: 8 quarters and 16 dimes
5. Last problem. Let's use substitution as well.
a-number of adult tickets
s-number of student tickets
a+s=530 ----> a=530-s
4a+3s=1740
4(530-s)+3s=1740
2120-4s+3s=1740
-s=-380
s=380
a+s=530
a+380=530
a=150
answer: 380 student tickets and 150 adult tickets
I hope this helps!
