QuanticaAlgebra IOpen in the courseEvaluating was settled in 1.6, so nothing about the method changes here, the expressions just get heavier. Substitute in parentheses, compute by the order of operations, read off one number. This lesson is six problems, and each one adds a step the last did not have.
Problem
Evaluate \(5-4y\) at \(y=-7\). There are two minus signs in the second term, one printed in the expression and one that comes with the value, so write the substitution with parentheses to keep them separate.
Show a hint
- Read \(4(-7)\) as one finished product before subtracting anything.
Show the full solution
Substituting in parentheses gives \(5-4(-7)\). The product is \(4(-7)=-28\), and subtracting \(-28\) is adding \(28\), so the value is \(5+28=\boxed{33}\). Losing the value's sign turns the line into \(5-28=-23\), which is exactly what the parentheses prevent.
Problem
Evaluate \(\dfrac{3k}{4}+\dfrac{20}{k}\) at \(k=8\). The variable sits under a fraction bar this time, and it appears in two places at once. Enter the value as a fraction in lowest terms.
Show a hint
- Both \(k\)s become \(8\) at the same moment, one above a bar and one below, and each bar finishes as its own division.
Show the full solution
Substituting gives \(\frac{3(8)}{4}+\frac{20}{8}\). Each bar finishes as its own division, \(\frac{24}{4}=6\) and \(\frac{20}{8}=\frac{5}{2}\), so the value is \(6+\frac{5}{2}=\boxed{\dfrac{17}{2}}\). Combining across the bars into \(\frac{24+20}{4+8}\) merges two separate divisions into one and is wrong.
Problem
Evaluate \(\sqrt{-12w}\) at \(w=-3\). The minus printed under the bar looks like the even-root trouble from 1.5. Substitute first, then decide whether the root exists.
Show a hint
- Compute the number actually under the bar once \(w\) is replaced, then judge.
Show the full solution
Substituting puts \(-12(-3)=36\) under the bar, and \(\sqrt{36}=\boxed{6}\). The minus printed in \(-12w\) is not the sign of the value, so rejecting the root on sight would throw away a perfectly good number.
Problem
Evaluate \(n^2-5n+3\) at \(n=-2\). Three terms, three sign decisions, one value, and every appearance of \(n\) gets its own parentheses.
Show a hint
- Write the parentheses at both appearances of \(n\) before doing any arithmetic.
- Subtracting \(5(-2)\) is subtracting a negative.
Show the full solution
Substituting at both appearances gives \((-2)^2-5(-2)+3\). That is \(4+10+3=\boxed{17}\). Two slips are common here, reading the square as \(-2^2=-4\), and dropping the sign in the middle term when \(-5(-2)\) is really \(+10\).
Problem
Evaluate \((2h-3)^2+h\) at \(h=6\). The parentheses hold a whole expression this time. Build the inside value first, square the result, then finish the sum.
Show a hint
- The exponent applies to the whole parenthesized value, so finish \(2h-3\) before squaring anything.
Show the full solution
Inside first, \(2(6)-3=9\). Squaring gives \(9^2=81\), and the total is \(81+6=\boxed{87}\). The exponent takes the finished inside value, so squaring term by term into \(144-9+6=141\) is wrong.
Problem
This one uses every skill from the lesson. Evaluate \(\dfrac{\sqrt{1-12v}}{v}\) at \(v=-4\). The bar covers all of \(1-12v\), the input is negative, and a division finishes the job. Enter the value as a fraction in lowest terms.
Show a hint
- Substitute before touching the root, so compute \(1-12(-4)\) under the bar first.
- After taking the root, divide by \(v\) itself. Its value is \(-4\), sign included.
Show the full solution
Under the bar, \(1-12(-4)=1+48=49\), so the top is \(\sqrt{49}=7\). The bottom is \(v=-4\), and \(7\div(-4)=\boxed{-\dfrac{7}{4}}\). Two slips lurk here, stopping at the radical because of the printed minus, and dropping the denominator's sign to land on \(\dfrac{7}{4}\).
The expressions got heavier and the method never changed. Parentheses on every substitution, signs judged after substituting, computation run from the inside out. Next, 2.3 turns the distributive property around, reading \(ab+ac\) as \(a(b+c)\), a rewrite called factoring.
Practice these ideas
Practice
Evaluate \(9-2t\) at \(t=-3\).
Show the solution
Substituting gives \(9-2(-3)\). The product is \(-6\), and subtracting \(-6\) is adding \(6\), so the value is \(9+6=\boxed{15}\). Without the parentheses the term reads \(2\cdot 3\) and the answer comes out \(3\).
Practice
Evaluate \(\dfrac{m}{3}+m\) at \(m=12\).
Show the solution
Only the first term sits over the bar, so it is \(12\div 3=4\), and the value is \(4+12=\boxed{16}\). The bar groups only its own top, so pulling the lone \(m\) underneath into \(\frac{12+12}{3}\) is wrong.
Practice
Evaluate \(\sqrt{2-7d}\) at \(d=-2\). Substitute before judging the root.
Show the solution
Substituting gives \(-7(-2)=14\), so the bar holds \(2+14=16\) and \(\sqrt{16}=\boxed{4}\). What decides whether an even root exists is the sign of the value after substituting, not the sign printed in the expression.
Practice
Evaluate \(w^2+3w\) at \(w=-18\).
Show the solution
Substituting gives \((-18)^2+3(-18)\), which is \(324-54=\boxed{270}\). The square takes the whole value with its sign, so \((-18)^2=324\). Reading it as \(-18^2=-324\) drops the total to \(-378\).
Practice
Evaluate \((3g-4)^2\) at \(g=5\).
Show the solution
Inside first, \(3(5)-4=11\), and \(11^2=\boxed{121}\). The exponent applies to the finished parenthesized value, so squaring \(g\) alone into \(3(25)-4=71\) is wrong.
Practice
Evaluate \(\dfrac{40}{p}-p\) at \(p=-5\).
Show the solution
The first term is \(40\div(-5)=-8\), and subtracting \(p\) means subtracting \(-5\), so the value is \(-8+5=\boxed{-3}\). The divisor and the subtracted \(p\) both keep their signs, and dropping either one lands you on \(13\).
Practice
Evaluate \(\sqrt{h^2+63}\) at \(h=-1\).
Show the solution
\((-1)^2=1\), so the bar holds \(1+63=64\), and \(\sqrt{64}=\boxed{8}\). Reading the square as \(-1^2=-1\) leaves \(62\) under the bar, which has no clean root, a good sign you took the sign wrong.
Practice
Evaluate \(\dfrac{a^2+b}{a+b}\) at \(a=-3\) and \(b=6\).
Show the solution
The top is \((-3)^2+6=15\) and the bottom is \(-3+6=3\), so the value is \(15\div 3=\boxed{5}\). Finish the whole top and the whole bottom before dividing, and remember \(a^2\) means \((-3)^2=9\), not \(-9\).
Practice
Evaluate \(2c^2-c^3\) at \(c=-3\).
Show the solution
The even power clears the sign, so \(2(-3)^2=18\), and the odd power keeps it, so \((-3)^3=-27\). Then \(18-(-27)=18+27=\boxed{45}\). Treating the cube as \(+27\) gives \(-9\) instead.
Practice
Evaluate \(\dfrac{\sqrt{50-t^2}}{t+12}\) at \(t=-5\). Enter the value as a fraction in lowest terms.
Show the solution
The bar holds \(50-(-5)^2=50-25=25\), so the top is \(5\). The bottom is \(-5+12=7\), and the value is \(\boxed{\dfrac{5}{7}}\). Both halves turn on signs, \((-5)^2\) is \(25\) rather than \(-25\), and the bottom is \(-5+12\) rather than \(-5-12\).