Here's how you solve this. So, x+y=2, right? Let's isolate x. x+y-y=2-y. x=2-y. NOW, if x=2-y, in 3x+2y=5, we can REPLACE x with 2-y and use it to solve for y! 3x+2y=5. 3(2-y)+2y=5. (3*2)+(3*-y)+2y=5. 6+(-3y)+2y=5. 6+(-y)=5. 6+(-y)-6=5-6. -y=-1. -y/-1=-1/-1. y=1.
So, if y=1, we can substitute that back into either equation--but let's go with the easier one, x+y=2. x+1=2. x+1-1=2-1. x=1. 1+1=2, so that works; let's check the other equation. 3(1)+2(1)=5. 3+2=5. 5=5. That's correct!
Answer: x=1, y=1
Answer:
(a) 
(b) Yaw, Musa and Kofi
Step-by-step explanation:
Given



Solving (a): The cost of the business
First, we solve for the fraction of Musa







So, we have:

Make Total the subject



Solving (b): Partners in ascending order
First, represent the fractions as decimals



From the conversion above, the least is 0.1333 (Yaw), then 0.3111 (Musa), then 0.5556 (Kofi).
<em>So, the order is: Yaw, Musa and Kofi</em>
Hello there.
6(8 - 2y) = 4y
To solve for this, we need to apply the Distributive Property to the left side of the equation. This property allows us to multiply the number outside of the parenthesis by all numbers inside of the parenthesis.
6(8 - 2y)
6(8) + 6(-2y)
48 - 12y
Now, let’s take a look at our equation.
-12y + 48 = 4y
To make things more simple, we’ll add 12y to both sides of the equation. This will cancel out -12y on the left side of the equation and will turn 4y on the right side of the equation into 16y.
Our new equation is:
16y = 48
Now all we need to do is divide both sides by 16 to solve for y.
16y / 16 = y
48 / 16 = 3
Our final answer and solution is:
Y = 3
I hope this helps!
Answer:
• David
,
• 4 miles
Explanation:
In the graph:
The given locations are:
• Owen's House, A(11,3)
,
• David's House, B(15,13)
,
• School, C(3,18)
We determine both Owen's and David's distance from the school using the distance formula.

Owen's distance from school (AC)
![\begin{gathered} AC=\sqrt[]{(3-11)^2+(18-3)^2} \\ =\sqrt[]{(-8)^2+(15)^2} \\ =\sqrt[]{64+225} \\ =\sqrt[]{289} \\ AC=17\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20AC%3D%5Csqrt%5B%5D%7B%283-11%29%5E2%2B%2818-3%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-8%29%5E2%2B%2815%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B64%2B225%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B289%7D%20%5C%5C%20AC%3D17%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
David's distance from school (BC)
![\begin{gathered} BC=\sqrt[]{(3-15)^2+(18-13)^2} \\ =\sqrt[]{(-12)^2+(5)^2} \\ =\sqrt[]{144+25} \\ =\sqrt[]{169} \\ BC=13\text{ miles} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20BC%3D%5Csqrt%5B%5D%7B%283-15%29%5E2%2B%2818-13%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B%28-12%29%5E2%2B%285%29%5E2%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B144%2B25%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B169%7D%20%5C%5C%20BC%3D13%5Ctext%7B%20miles%7D%20%5Cend%7Bgathered%7D)
We see from the calculations that David lives closer to the school, and by 4 miles.
The graph below is attached for further understanding:
Answer:
x=12
Step-by-step explanation:
Isolate the variable by dividing each side by factors that don't contain the variable.