Sample Checkpoint
Checkpoint 6A
Problem 6-29
Rewriting Equations With More Than One Variable
Answers to problem 6-29: a: y=3x−105=35x−2, b: x=y−bm, c: r2=Aπ
Rewriting equations with more than one variable may be done in a variety of ways but normally you follow the same steps as you would for solving an equation with a single variable. Commonly, the first step is to multiply all the terms by a common denominator to remove all of the fractions. Then solve in the usual way. Collect like terms. Isolate the specified variable terms on one side and everything else on the other. Finally, divide or undo the exponent so that the variable is alone. The final answer will be an equation that involves variables and possibly numbers.
Example 1: Solve for y: 2x+3y−9=0
Solution: 2x + 3y − 9 = 0
problem
3y=−2x+9
subtract 2x, add 9 on each side
y=−2x+93
3 divide by 3
y=−23x+3
simplify
Example 2: Solve for r: V = 43πr3
Solution: V = 43πr3
problem
3V = 4πr3
multiply by 3 on each side
3V4π=r3
divide by 4π
3V4π3=r
cube root
Now we can go back and solve the original problems.
−3x+5y=−105y=3x−10 y=3x−105 y=x−2
y=mx+by−b=mx+by−bm=x
A=πr2Aπ=r2
Here are some more to try. Solve each equation for the specified variable.
2x−3y=9(for x)
2x−3y=9(for y)
5x+3y=15(for y)
4n=3m−1(for m)
2w+2l=P(for w)
2a+b=c (for a)
I=ER(for R)
y=14x+1(for x)
c2=a2+b2(for b2)
V=s3(for s)
S=4πr2(for r2)
m=a+b+c3(for a)
a3+b=c2(for a)
Fa=m(for a)
A=12h(b1+b2)(for b1)
V=13πr2h(for h)
Answers:
x=3y+92
y=2x−93
y=15−5×3
m=4n+13
w=P−2l2
a=c−b2
R=EI
x=4(y−1)
b2=c2−a2
s=V3
r2=S4π
a=3m−b−c
a=c2−b3
a=Fm
b1=2Ah−b2
h=3Vπr2