12 multiple-choice questions, progressively harder.
Solve ax+b=cx+dax + b = cx + dax+b=cx+d for xxx and state the correct condition.
Solution
Correct answer: A
Collect and factor, then divide by the factor a−ca - ca−c.
(a−c)x=d−b ⇒ x=d−ba−c,a≠c(a - c)x = d - b \;\Rightarrow\; x = \frac{d - b}{a - c}, \quad a \neq c(a−c)x=d−b⇒x=a−cd−b,a=c
The condition is on the factor a−ca - ca−c, not on aaa and ccc separately.
Solve I=ER+rI = \dfrac{E}{R + r}I=R+rE for RRR (current from a source with internal resistance rrr).
Correct answer: D
Clear the denominator, then isolate RRR.
I(R+r)=E ⇒ IR=E−Ir ⇒ R=E−IrII(R + r) = E \;\Rightarrow\; IR = E - Ir \;\Rightarrow\; R = \frac{E - Ir}{I}I(R+r)=E⇒IR=E−Ir⇒R=IE−Ir
The formula R2=RR1R1−RR_2 = \dfrac{R R_1}{R_1 - R}R2=R1−RRR1 requires which condition?
Correct answer: C
The divisor is the factor R1−RR_1 - RR1−R.
R1−R≠0 ⇒ R1≠RR_1 - R \neq 0 \;\Rightarrow\; R_1 \neq RR1−R=0⇒R1=R
A parallel combination is always smaller than either resistor, so this holds automatically for a real circuit.
Solving ax=bax = bax=b for bbb requires which condition?
Correct answer: B
Solving for bbb is just b=axb = axb=ax, which involves no division.
b=ax is defined for all a,xb = ax \text{ is defined for all } a, xb=ax is defined for all a,x
A condition appears only when you divide by something that can vanish, and here nothing was divided.
Solve the infinite geometric sum S=a1−rS = \dfrac{a}{1 - r}S=1−ra for rrr, with its condition.
Clear the denominator and solve for rrr.
S(1−r)=a ⇒ S−Sr=a ⇒ r=S−aSS(1 - r) = a \;\Rightarrow\; S - Sr = a \;\Rightarrow\; r = \frac{S - a}{S}S(1−r)=a⇒S−Sr=a⇒r=SS−a
The last step divides by SSS, so the condition introduced by this rearrangement is S≠0S \neq 0S=0.
For which value of kkk does 3x+k=kx+33x + k = kx + 33x+k=kx+3 have every real number as a solution?
Reduce the equation.
3x−kx=3−k ⇒ (3−k)x=3−k3x - kx = 3 - k \;\Rightarrow\; (3 - k)x = 3 - k3x−kx=3−k⇒(3−k)x=3−k
Both sides carry the factor 3−k3 - k3−k, which vanishes at k=3k = 3k=3, giving 0⋅x=00 \cdot x = 00⋅x=0: every real number.
Solve 1p+1q=1r\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{1}{r}p1+q1=r1 for qqq.
Isolate 1q\frac{1}{q}q1, combine over a common denominator, then take reciprocals.
1q=1r−1p=p−rpr ⇒ q=prp−r\frac{1}{q} = \frac{1}{r} - \frac{1}{p} = \frac{p - r}{pr} \;\Rightarrow\; q = \frac{pr}{p - r}q1=r1−p1=prp−r⇒q=p−rpr
Solve a(x+1)=b(x−1)a(x + 1) = b(x - 1)a(x+1)=b(x−1) for xxx, with the correct condition.
Expand, collect the xxx terms, factor, and divide.
ax+a=bx−b ⇒ (a−b)x=−(a+b) ⇒ x=a+bb−aax + a = bx - b \;\Rightarrow\; (a - b)x = -(a + b) \;\Rightarrow\; x = \frac{a + b}{b - a}ax+a=bx−b⇒(a−b)x=−(a+b)⇒x=b−aa+b
Dividing by the factor a−ba - ba−b needs a≠ba \neq ba=b.
Solve ax−b=bx−aax - b = bx - aax−b=bx−a for xxx, assuming a≠ba \neq ba=b.
Collect the xxx terms and factor.
ax−bx=b−a ⇒ (a−b)x=−(a−b)ax - bx = b - a \;\Rightarrow\; (a - b)x = -(a - b)ax−bx=b−a⇒(a−b)x=−(a−b)
Divide by a−ba - ba−b (nonzero since a≠ba \neq ba=b): x=−1x = -1x=−1, the same value for every such aaa and bbb.
Solve P=2(l+w)P = 2(l + w)P=2(l+w) for www.
Distribute, subtract 2l2l2l, then divide the whole side by 222.
P=2l+2w ⇒ w=P−2l2P = 2l + 2w \;\Rightarrow\; w = \frac{P - 2l}{2}P=2l+2w⇒w=2P−2l
This equals P2−l\frac{P}{2} - l2P−l, not P2−2l\frac{P}{2} - 2l2P−2l.
A family of equations reduces to (k−8)x=k−4(k - 8)x = k - 4(k−8)x=k−4. For which value of kkk is there no solution?
No solution needs the coefficient zero and the constant nonzero.
k−8=0 ⇒ k=8,k−4=4≠0k - 8 = 0 \;\Rightarrow\; k = 8, \qquad k - 4 = 4 \neq 0k−8=0⇒k=8,k−4=4=0
At k=8k = 8k=8 the equation is 0⋅x=40 \cdot x = 40⋅x=4, so there is no solution.
Solve the point-slope equation y−y1=m(x−x1)y - y_1 = m(x - x_1)y−y1=m(x−x1) for xxx (with m≠0m \neq 0m=0).
Divide both sides by mmm, then add x1x_1x1.
y−y1m=x−x1 ⇒ x=x1+y−y1m\frac{y - y_1}{m} = x - x_1 \;\Rightarrow\; x = x_1 + \frac{y - y_1}{m}my−y1=x−x1⇒x=x1+my−y1
Valid since m≠0m \neq 0m=0.
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