Answer:
x= 12 and y= 2
Step-by-step explanation:
First you would line up the equations so the x's and y's are on top of each other. Then you would multiply x+y=14 by 3 to give you 3x+3y=42. After that you subtract 2x-3y=30 and 3x+3y=42 to give you an answer of x=12. After that, you plug in x with 12 in the equation x+y=14. You subtract 12 from both sides to get an answer of 2. So ur solution is (12,2)
Answer:
is this the answer im sorry i did not put steps i was in a hurry to go to the next class.
Step-by-step explanation:
y=10(log(5x)+1)/9lq(y,x)
l=0
Answer: 2 and 3
Step-by-step explanation:
I did this and got it right
You're looking for the largest number <em>x</em> such that
<em>x</em> ≡ 1 (mod 451)
<em>x</em> ≡ 4 (mod 328)
<em>x</em> ≡ 1 (mod 673)
Recall that
<em>x</em> ≡ <em>a</em> (mod <em>m</em>)
<em>x</em> ≡ <em>b</em> (mod <em>n</em>)
is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly
1 ≡ 4 (mod 41)
is not true.
So there is no such number.
