16*8=128 and 128*12= is 1,536
A)
<span>|x + y = 5 </span>
<span>|2x - y = 7; </span>
<span>b) </span>
<span>|2x + y = 5 </span>
<span>|x - y = 2 </span>
<span>c) </span>
<span>|3x + y = 6 </span>
<span>|4x - 3y = -5 </span>
<span>d) </span>
<span>|1/(x - 1) = y - 3 </span>
<span>|x - y = -2 </span>
<span>e) </span>
<span>|(9x + 4y)/3 - (5x - 11)/2 = 13 - y </span>
<span>|13x - 7y = -8 </span>
<span>Answer: </span>c<span> and </span>e<span> has solution (1; 3)</span>
Answer:
13000
Step-by-step explanation:
You can use systems of equations for this one.
We are going to use 'q' as the number of quarters Rafael had,
and 'n' as the number of nickels Rafael had.
You can write the first equation like this:
3.50=0.05n+0.25q
This says that however many 5 cent nickels he had, and however many
25 cent quarters he had, all added up to value $3.50.
Our second equation is this:
q=n+8
This says that Rafael had 8 more nickels that he had quarters.
We can now use substitution to solve our system.
We can rewrite our first equation from:
3.50=0.05n+0.25q
to:
3.50=0.05n+0.25(n+8)
From here, simply solve using PEMDAS.
3.50=0.05n+0.25(n+8) --Distribute 0.25 to the n and the 8
3.50=0.05n+0.25n+2 --Subtract 2 from both sides
1.50=0.05n+0.25n --Combine like terms
1.50=0.30n --Divide both sides by 0.30
5=n --This is how many NICKELS Rafael has.
We now know how many nickels he has, but the question is asking us
how many quarters he has.
Simply substitute our now-known value of n into either of our previous
equations (3.50=0.05n+0.25q or q=n+8) and solve.
We now know that Rafael had 13 quarters.
To check, just substitute our known values for our variables and solve.
If both sides of our equations are equal, then you know that you have
yourself a correct answer.
Happy math-ing :)
