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Literal Equations and Formulas: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Reading a sensor

The real quantities satisfy $R=\frac{3a-2x}{5}$ . Solve for $a$ , treating $R$ and $x$ as parameters.
1. Hint 1
  The requested unknown is the letter that must stand alone.
2. Hint 2
  Clear the denominator, then move the term containing $x$ to the other side.
Answer

$a=\frac{5R+2x}{3}$ ; no new restrictions.

Full solution

The unknown is $a$ and the fixed parameters are $R$ and $x$ .
$\begin{aligned}5R&=3a-2x\\3a&=5R+2x\\a&=\frac{5R+2x}{3}\end{aligned}$

The divisor $3$ is nonzero, so there is no new restriction.
Substitution makes the original numerator $5R$ .

Answer

$a=\frac{5R+2x}{3}$ ; no new restrictions.

Key idea

A letter becomes the unknown because the question asks to isolate it.
Problem 1: Mark complete
Problem 2 A combined reading

Solve $M=pu+2qu-v$ for $u$ , assuming the real parameters satisfy $p+2q\ne0$ .
1. Hint 1
  The two terms containing $u$ need to be collected together.
2. Hint 2
  Add $v$ and factor $u$ out of $pu+2qu$ .
Answer

$u=\frac{M+v}{p+2q}$ .

Full solution

Collect the unknown and divide by the stated nonzero factor.
$\begin{aligned}M+v&=u(p+2q)\\u&=\frac{M+v}{p+2q}\end{aligned}$

Substituting makes $pu+2qu=M+v$ , so the original right side is $M$ .

Answer

$u=\frac{M+v}{p+2q}$ .

Key idea

The divisor is the whole coefficient of the collected unknown.
Problem 2: Mark complete
Problem 3 A remaining share

The formula $B=A-r(A-C)$ involves real numbers. Solve for $r$ when $A\ne C$ .
1. Hint 1
  Treat $A-C$ as one fixed factor.
2. Hint 2
  Subtract $B$ from $A$ to isolate the product containing $r$ .
Answer

$r=\frac{A-B}{A-C}$ .

Full solution

Rearrange without splitting the coefficient.
$\begin{aligned}A-B&=r(A-C)\\r&=\frac{A-B}{A-C}\end{aligned}$

The division is allowed by $A\ne C$ .
Substituting replaces $r(A-C)$ with $A-B$ and recovers $B$ .

Answer

$r=\frac{A-B}{A-C}$ .

Key idea

A grouped difference acts as a single coefficient when solving for another letter.
Problem 3: Mark complete
Problem 4 A reservoir setting

A reservoir starts with $B$ liters. Water enters at $p$ liters per minute and leaves at $q$ liters per minute, where $p>q\ge0$ . The desired amount is $T\ge B$ . Find the required time $t$ in minutes and explain why the formula gives an allowed time.
1. Hint 1
  The amount changes at the net rate of entry minus exit.
2. Hint 2
  Write the final amount as the starting amount plus net rate times time.
Answer

$t=\frac{T-B}{p-q}$ minutes; $t\ge0$ .

Full solution

The amount after $t$ minutes is $B+(p-q)t$ .
$\begin{aligned}T-B&=(p-q)t\\t&=\frac{T-B}{p-q}\end{aligned}$

The denominator is positive because $p>q$ , and the numerator is nonnegative because $T\ge B$ .
Thus the time is nonnegative.
Substitution gives the desired amount $T$ .

Answer

$t=\frac{T-B}{p-q}$ minutes; $t\ge0$ .

Key idea

Context can guarantee both a nonzero divisor and an admissible value of the unknown.
Problem 4: Mark complete
Problem 5 A weighted mixture

A mixture of $u$ liters at concentration $a$ and $v$ liters at concentration $b$ has concentration $c$ , so $c(u+v)=au+bv$ . Here $v>0$ and $a>c>b\ge0$ . Find $u$ in factored form and check that it is positive.
1. Hint 1
  The unknown appears on both sides of the concentration equation.
2. Hint 2
  Collect the terms containing $u$ and keep the differences of concentrations grouped.
Answer

$u=\frac{v(c-b)}{a-c}$ liters; $u>0$ .

Full solution

Expand the left side and collect.
$\begin{aligned}cu+cv&=au+bv\\v(c-b)&=u(a-c)\\u&=\frac{v(c-b)}{a-c}\end{aligned}$

The divisor $a-c$ is positive, and both $v$ and $c-b$ are positive.
Thus $u>0$ , as a volume must be.
Rearranging the displayed product back verifies the mixture equation.

Answer

$u=\frac{v(c-b)}{a-c}$ liters; $u>0$ .

Key idea

Collecting a repeated unknown can make the physical conditions on a formula explicit.
Problem 5: Mark complete
Problem 6 An adjustable record

For real parameters $k$ and $d$ , solve $k(x+2)=4x+d$ completely for real $x$ , including every case where there is no single solution.
1. Hint 1
  Reduce the equation to one coefficient times $x$ .
2. Hint 2
  When that coefficient vanishes, evaluate the remaining constant instead of dividing.
Answer

$x=\frac{d-2k}{k-4}$ if $k\ne4$ ; all real $x$ if $k=4,d=8$ ; no solution if $k=4,d\ne8$ .

Full solution

Expand and collect the unknown.
$\begin{aligned}kx+2k&=4x+d\\(k-4)x&=d-2k\end{aligned}$

If $k\ne4$ , division gives
$x=\frac{d-2k}{k-4}.$
If $k=4$ , the original reduces to $8=d$ , which holds for every $x$ when $d=8$ and for no $x$ otherwise.

Answer

$x=\frac{d-2k}{k-4}$ if $k\ne4$ ; all real $x$ if $k=4,d=8$ ; no solution if $k=4,d\ne8$ .

Key idea

The constant remaining when a parameter coefficient vanishes decides between all solutions and none.
Problem 6: Mark complete
Problem 7 A calibration fraction

A calibration formula is $s=\frac{z-r}{z+r}$ , where $r\ne0$ is a fixed real parameter and $z\ne-r$ . Solve for $z$ for each real value of $s$ , and retain the original restriction.
1. Hint 1
  Keep the original excluded value before clearing the fraction.
2. Hint 2
  Collect both terms containing $z$ after multiplying by $z+r$ .
Answer

$z=\frac{r(1+s)}{1-s}$ if $s\ne1$ ; no solution if $s=1$ ; the formula never gives $z=-r$ .

Full solution

Because $z\ne-r$ , multiplying by $z+r$ is reversible.
$\begin{aligned}sz+sr&=z-r\\(s-1)z&=-r(1+s)\end{aligned}$

For $s\ne1$ , this gives
$z=\frac{r(1+s)}{1-s}.$
Its sum with $r$ is $\frac{2r}{1-s}$ , nonzero, so it respects $z\ne-r$ .
At $s=1$ , the equation becomes $r=-r$ , impossible because $r\ne0$ .

Answer

$z=\frac{r(1+s)}{1-s}$ if $s\ne1$ ; no solution if $s=1$ ; the formula never gives $z=-r$ .

Key idea

A rearranged fraction must satisfy both the new divisor condition and the original domain.
Problem 7: Mark complete
Problem 8 Three parameter choices

A student says $au+bu=c$ has a unique solution whenever at least one of $a$ and $b$ is nonzero. Decide whether this is correct, and give parameter values that support your conclusion.
1. Hint 1
  Uniqueness depends on the coefficient after collecting the unknown.
2. Hint 2
  Two nonzero terms may cancel when their coefficients are added.
Answer

False; for example $a=1,b=-1,c=2$ gives no solution.

Full solution

Factoring gives
$(a+b)u=c.$

The required condition for uniqueness is $a+b\ne0$ .
Choosing $a=1,b=-1,c=2$ makes the equation $0u=2$ , which has no solution although both coefficients are nonzero.

Answer

False; for example $a=1,b=-1,c=2$ gives no solution.

Key idea

Nonzero individual terms do not guarantee a nonzero combined coefficient.
Problem 8: Mark complete
Problem 9 Two adjustable coefficients

Choose real numbers $a$ and $b$ so that $a(x+2)=b(x-1)+9$ is true for every real $x$ . Starting with your chosen pair, can changing exactly one of $a$ and $b$ make the equation have no solution? Justify your answer.
1. Hint 1
  An identity needs the coefficients of $x$ and the constant terms to agree.
2. Hint 2
  After finding the pair, consider what changing only one coefficient does to the coefficient of $x$ in the reduced equation.
Answer

$a=b=3$ ; changing exactly one of them cannot produce no solution.

Full solution

Collect the terms containing $x$ .
$(a-b)x=9-2a-b$
For an identity, both sides must have zero coefficient or constant, so $a=b$ and $9-2a-b=0$ .
Substitution gives $9-3a=0$ , hence $a=b=3$ .

Check the pair: both original sides expand to $3x+6$ .
Changing exactly one of $a,b$ makes them unequal.
The coefficient $a-b$ is then nonzero, so the changed equation has exactly one solution, rather than no solution.

Answer

$a=b=3$ ; changing exactly one of them cannot produce no solution.

Key idea

Changing one coefficient can prevent the zero-coefficient case that a contradiction requires.
Problem 9: Mark complete
Problem 10 Two requests for a formula

The relation $m=at+b$ is given with real quantities. One person solves for $a$ ; another solves for $b$ . Explain why the first request can need cases while the second does not, and give both complete answers.
1. Hint 1
  Changing the chosen unknown changes which quantities are treated as fixed.
2. Hint 2
  Isolating $a$ involves division by $t$ ; isolating $b$ involves subtraction.
Answer

$a=\frac{m-b}{t}$ if $t\ne0$ ; all real $a$ if $t=0,m=b$ ; none if $t=0,m\ne b$ . Always $b=m-at$ .

Full solution

For the first request, $m,t,b$ are parameters.
$at=m-b$

If $t\ne0$ , divide by $t$ .
At $t=0$ , this becomes $0=m-b$ , giving every $a$ when $m=b$ and none otherwise.
For the second request, subtract $at$ to get $b=m-at$ ; this needs no division and is valid for all real parameter values.

Answer

$a=\frac{m-b}{t}$ if $t\ne0$ ; all real $a$ if $t=0,m=b$ ; none if $t=0,m\ne b$ . Always $b=m-at$ .

Key idea

The same relation can require a case split for one unknown and a single formula for another.
Problem 10: Mark complete

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