JKM = M + L
15x - 48 = 40 + 5x + 12
10x - 48 = 52
10x = 100
x = 10
JKM = 15(10) - 48 = 102
180 = JKM + MKL
180 = 102 + MKL
MKL = 180 - 102 = 78
The solution to given system of equations is (x, y) = (2, -1)
<em><u>Solution:</u></em>
Given system of equations are:
-1x + 2y = -4 -------- eqn 1
4x + 3y = 5 ------- eqn 2
We can solve the above system of equations by elimination method
<em><u>Multiply eqn 1 by 4</u></em>
4(-1x + 2y = -4)
-4x + 8y = -16 ------ eqn 3
<em><u>Add eqn 2 and eqn 3</u></em>
4x + 3y = 5
-4x + 8y = -16
( + ) --------------------
0x + 11y = -16 + 5
11y = -11
y = -1
<em><u>Substitute y = -1 in eqn 1</u></em>
-1x + 2(-1) = -4
-x -2 = -4
-x = -4 + 2
-x = -2
x = 2
<em><u>Check the answer:</u></em>
Substitute x = 2 and y = -1 in eqn 2
4x + 3y = 5
4(2) + 3(-1) = 5
8 - 3 = 5
5 = 5
Thus the obtained answer is correct
Thus the solution to given system of equations is (x, y) = (2, -1)
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.
Answer:
x/23
Step-by-step explanation:
The bag of money can be represented by x
Adam is 1 person.
1 + 22 = 23
x is divided by 23
x/23
~ Zoe
Sarah starts with 3 fish
in 1 month she will have 6 fish
after 2 months she will have 12
in 3 months she will have 24
