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Answer to Question #14469 in Algebra for Jaclyn
Question #14469
Make the variable in the bracket the subject of the given formula:
1. px-q=rx+s {x}
2. s = a/(1-r) {r}
3. A = 1/2(a+b)h {a}
1. px-q=rx+s {x}
2. s = a/(1-r) {r}
3. A = 1/2(a+b)h {a}
Expert's answer
1. px-q=rx+s {x}
px-rx=s+q
x(p-r)=s+q
x=(s+q)/(p-r)
2. s = a/(1-r) {r}
s(1-r)=a
s-sr=a
sr=s-a
r=(s-a)/s
3. A = 1/2{(a+b)h} {a}
2A=ah+bh
ah=2A-bh
a=(2A-bh)/h
px-rx=s+q
x(p-r)=s+q
x=(s+q)/(p-r)
2. s = a/(1-r) {r}
s(1-r)=a
s-sr=a
sr=s-a
r=(s-a)/s
3. A = 1/2{(a+b)h} {a}
2A=ah+bh
ah=2A-bh
a=(2A-bh)/h
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#340153 on Dec 2023
#340153 on Dec 2023
Comments
ab/{a+b} =1 ab=a+b ab-a=b a(b-1)=b a=b/(b-1)
Then ab/a+b = 1 {a}
Ok, then A = 1/{2(a+b)h} whence 2(a+b)h=1/A a+b=1/{2hA} a=1/{2hA}-b
2{(a+b)h} is a denominator
A = 1/2{(a+b)h}=1/2*{(a+b)h} so 2*A={(a+b)h} Or you've meant that 2{(a+b)h} is a denominator?
3. A = 1/2{(a+b)h} {a} 2A=ah+bh ah=2A-bh a=(2A-bh)/h How did you get 2A?
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