Answer:
(0, -7) and (5, 0)
Step-by-step explanation:
y-intercept is when x = 0, therefore (0, -7)
x-intercept is when y = 0, therefore (5, 0)
We have 3 people: Paul, Daniel and Patricia. Let each name denote that person’s current age in years. Remember that % is the symbol for 1/100.
PAUL: (#A) Paul = Patricia – 2
DANIEL:
(#B) Daniel = 125% * Patricia = 125 * (1/100) * Patricia = 1.25 * Patricia
(#C) Ten years ago, Daniel’s age at that time would have been Daniel’s current age – 10. IE “Daniel – 10”.
Ten years ago, Patricia’s age at that time would have been Patricia’s current age – 10. IE “Patricia – 10”.
Ten years ago, Daniel’s age at that time (IE Daniel – 10) was 50% greater than Patricia’s age at the time (Patricia – 10).
IE
Daniel – 10 = 150% * (Patricia – 10)
Daniel – 10 = 1.50 * (Patricia – 10)
Daniel – 10 = 1.50 * Patricia – 1.50 * 10
Daniel – 10 = 1.50 * Patricia – 15
Daniel = 1.50 * Patricia – 15 + 10
Daniel = 1.50 * Patricia – 5
So now we have 3 equations in 3 unknowns:
(A) Paul = Patricia – 2
(B) Daniel = 1.25 * Patricia
(C) Daniel = 1.50 * Patricia – 5
Solve equations B & C simultaneously
1.25 * Patricia = 1.50 * Patricia – 5
1.25 * Patricia – 1.50 * Patricia = – 5
– 0.25 * Patricia = – 5
Patricia = (– 5) / (– 0.25)
Patricia = + 20
So now we can write the 3 people’s ages:
(a) Patricia = 20
(b) Paul = Patricia – 2 = 20 – 2 = 18
(c) Daniel = 1.25 * Patricia = 1.25 * 20 = 25
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<span>Finally, we check our solution against the original question. </span>
Paul is two years younger than Patricia. ……… 20 – 2 = 18 ✔
Daniel is 25% older than Patricia. ……… 25 = 125% * 20 ✔
Ten years ago, Daniel was 50% older than Patricia. 25 – 10 = 150% (20– 10) ✔
Hope that helped :)
Answer:
I'm not 100% sure but if I had to pick one I would go with B.
Step-by-step explanation:
Let me know if I'm correct if you decide to pick mine
well, we know it's a rectangle, so that means the sides JK = IL and JI = KL, so
![\stackrel{JK}{3x+21}~~ = ~~\stackrel{IL}{6y}\implies 3(x+7)=6y\implies x+7=\cfrac{6y}{3} \\\\\\ x+7=2y\implies \boxed{x=2y-7} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{JI}{6y-6}~~ = ~~\stackrel{KL}{2x+20}\implies 6(y-1)=2(x+10)\implies \cfrac{6(y-1)}{2}=x+10 \\\\\\ 3(y-1)=x+10\implies 3y-3=x+10\implies \stackrel{\textit{substituting from the 1st equation}}{3y-3=(2y-7)+10} \\\\\\ 3y-3=2y+3\implies y-3=3\implies \blacksquare~~ y=6 ~~\blacksquare ~\hfill \blacksquare~~ \stackrel{2(6)~~ - ~~7}{x=5} ~~\blacksquare](https://tex.z-dn.net/?f=%5Cstackrel%7BJK%7D%7B3x%2B21%7D~~%20%3D%20~~%5Cstackrel%7BIL%7D%7B6y%7D%5Cimplies%203%28x%2B7%29%3D6y%5Cimplies%20x%2B7%3D%5Ccfrac%7B6y%7D%7B3%7D%20%5C%5C%5C%5C%5C%5C%20x%2B7%3D2y%5Cimplies%20%5Cboxed%7Bx%3D2y-7%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7BJI%7D%7B6y-6%7D~~%20%3D%20~~%5Cstackrel%7BKL%7D%7B2x%2B20%7D%5Cimplies%206%28y-1%29%3D2%28x%2B10%29%5Cimplies%20%5Ccfrac%7B6%28y-1%29%7D%7B2%7D%3Dx%2B10%20%5C%5C%5C%5C%5C%5C%203%28y-1%29%3Dx%2B10%5Cimplies%203y-3%3Dx%2B10%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bsubstituting%20from%20the%201st%20equation%7D%7D%7B3y-3%3D%282y-7%29%2B10%7D%20%5C%5C%5C%5C%5C%5C%203y-3%3D2y%2B3%5Cimplies%20y-3%3D3%5Cimplies%20%5Cblacksquare~~%20y%3D6%20~~%5Cblacksquare%20~%5Chfill%20%5Cblacksquare~~%20%5Cstackrel%7B2%286%29~~%20-%20~~7%7D%7Bx%3D5%7D%20~~%5Cblacksquare)
Answer:
See picture:)
Hope that makes sense! If you'd like any more help with maths, I'd be happy to offer online tuition. You can find me at: www.birchwoodtutors.com
