Answer:
x = 3 and y = 2
or
(3,2)
Step-by-step explanation:
first equation (LCM = 3 x 4 x 5 = 60)
(x+2)/5 - (9-x)/3 = (y - 6)/4
12(x+2) - 20(9-x) = 15(y - 6)
12x + 24 - 180 + 20x = 15y - 90
32x - 15y = -90 - 24 + 180
32x - 15y = 66
second equation:
8(x + 2) - (2x + y) = 2(x + 13)
8x + 16 - 2x - y = 2x + 26
6x - y + 16 = 2x + 26
4x - y = 10
Now you have 2 equations:
4x - y = 10 ---> y = 4x - 10
32x - 15y = 66
Substitute y = 4x - 10 into 32x - 15y = 66
32x - 15(4x - 10) = 66
32x - 60x + 150 = 66
-28x = -84
x = 3
y = 4(3) - 10
y = 12 - 10
y = 2
Solutions: x = 3 and y = 2
2x^2+11x-15 Is the answer to 7
Step-by-step explanation:
Answer:
1260
Step-by-step explanation:
180 x 2
Answer: 31.25 ft
Step-by-step explanation:
50 / 32 = x / 20
32x = 50(20)
x = 31.25 ft
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.
