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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## Comments

Assignment Expert07.09.12, 18:56ab/{a+b} =1 ab=a+b ab-a=b a(b-1)=b a=b/(b-1)

Jaclyn07.09.12, 18:52Then ab/a+b = 1 {a}

Assignment Expert07.09.12, 18:49Ok, then A = 1/{2(a+b)h} whence 2(a+b)h=1/A a+b=1/{2hA} a=1/{2hA}-b

Jaclyn07.09.12, 18:452{(a+b)h} is a denominator

Assignment Expert07.09.12, 18:43A = 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?

Jaclyn07.09.12, 18:363. 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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