Exercise 100 Page 341 - 3. Section 6.3 - Core Connections Integrated I, 2013

a To solve this equation for

b,

we have to isolate it on one side of the equation by subtracting

m x

from both sides of the equation.

y = m x + b

$LHS - m x = RHS - m x$

y - m x = b

Rearrange equation

b = y - m x

b Now we want to solve the same equation as in Part A, but this time for

x .

y = m x + b

Subtract terms

y - b = m x

Rearrange equation

m x = y - b

$LHS / m = RHS / m$

x = \frac{y - b}{m}

c To solve this equation for

t,

we can divide the whole equation by

p r .

I = p r t

$LHS / p r = RHS / p r$

\frac{I}{p r} = t

Rearrange equation

t = \frac{I}{p r}

d To solve this equation for

t,

we will isolate

p r t

on one side first. Then we can divide the whole equation by

p r

to isolate

t .

A = p + p r t

$LHS - p = RHS - p$

A - p = p r t

Rearrange equation

p r t = A - p

$LHS / p r = RHS / p r$

t = \frac{A - p}{p r}