1.9 Equations and an initial look at inequalities

Quantitative problems often state that two expressions represent the same quantity and ask which input makes that statement true.

Equation. A statement that two expressions have equal values.

An equation states that two expressions are equal. Solving means finding every value that makes the original equation true. Whatever operation is applied to one side must also be applied to the other. A solution is checked by substituting it into the original statement. If simplification produces an identity such as \(0=0\), all values in the original domain solve the equation. If it produces a contradiction such as \(0=5,\) there is no solution.

Linear equations are solved by collecting variable terms on one side and constants on the other. Fractions may be cleared by multiplying every term by the least common denominator. Any values excluded by an original variable denominator remain excluded. The same balancing principle applies to inequalities, except that multiplication or division by a negative number reverses the comparison.

1.9.1 Checking and classifying results

Check a proposed solution by substituting it into both sides of the original statement. An identity such as \(0=0\) means that every value in the original domain solves the equation. A contradiction such as \(0=5\) means that no value solves it. For equations with variable denominators, state restrictions before clearing fractions because multiplication can conceal a forbidden input.

For \(2x-7=3x+2\), subtracting \(2x\) and then 2 gives \(x=-9\). Substitution confirms that both sides equal \(-25\).

Worked Example: Solving and checking an equation

Solve \(2x-7=3x+2\). Subtracting \(2x\) and then 2 gives \(x=-9\). The check is

\[2(-9)-7=-25\quad\text{and}\quad3(-9)+2=-25.\]

Try It. Solve and check \(5x+3=2x+18\).
Check Your Work \(3x=15\), so \(x=5\). Both sides equal 28.

Worked Example: Finding a chemical-feed setting

A feed system delivers 18 L before a variable operating period and then adds 6 L for each hour \(t\). If the required total is 54 L,

\[18+6t=54,\]

so \(6t=36\) and \(t=6\) h. Substitution gives \(18+6(6)=54\) L.

Try It. Solve \(25+8t=73\) and interpret \(t\) as hours.
Check Your Work \(8t=48\), so \(t=6\) h.

Applied Problem: Scheduling tank filling. A tank already contains 12,500 L. Water enters at 850 L/min while 150 L/min leaves through a drain. How many minutes are required for the tank to contain 30,000 L?

Check Your Work The net inflow is \(850-150=700\) L/min. Solve \(12{,}500+700t=30{,}000\). Then \(700t=17{,}500\), so \(t=25\) min.

Worked Example: Reversing an inequality

Solve \(-3x+6>15\). Subtract 6 to obtain \(-3x>9\). Dividing by \(-3\) reverses the symbol, so \(x<-3\).

Try It. Solve \(-4x+1\le13\).
Check Your Work \(-4x\le12\), so \(x\ge-3\).

Worked Example: Clearing fractions in a detention-time equation

Solve \(t/3+t/6=10\). Multiplying every term by 6 gives

\[2t+t=60,\]

so \(3t=60\) and \(t=20\) min. The original left side is \(20/3+20/6=10\).

Try It. Solve \(q/4+q/12=16\).
Check Your Work Multiplying by 12 gives \(3q+q=192\), so \(q=48\).

1.9.2 Practice Problems

  1. Solve \(4x-7=13\).
    Check Your Work \(x=5\).
  2. Solve \(3(x-2)=15\).
    Check Your Work \(x=7\).
  3. Determine whether \(x=4\) solves \(2x+1=9\).
    Check Your Work Yes.
  4. Solve \(2x+5<11\).
    Check Your Work \(x<3\).
  5. Solve \(-5x\ge20\).
    Check Your Work \(x\le-4\).
  6. A tank contains 8,000 L and fills at 600 L/min. Solve \(8000+600t=20{,}000\) for the filling time.
    Check Your Work \(600t=12{,}000\), so \(t=20\) min.
  7. A dosing system must deliver at least 90 L. It has delivered 30 L and continues at 12 L/min. Solve \(30+12t\ge90\).
    Check Your Work \(12t\ge60\), so \(t\ge5\) min.
  8. A process equation is \(q/5+q/10=18\). Find \(q\).
    Check Your Work Multiplying by 10 gives \(2q+q=180\), so \(q=60\).