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Practice Problems

\(\textbf{1)}\) \(\text{Solve for }k. \,\, A=\frac{1}{2}h(b+k)\)

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The answer is \(k=\displaystyle\frac{2A}{h}-b, \,\,\, h\ne 0 \)

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\(\,\,\,\,\,A=\frac{1}{2}h(b+k)\)

\(\,\,\,\,\,2A=h(b+k)\)

\(\,\,\,\,\,\displaystyle\frac{2A}{h}=(b+k) \,\,\, h\ne 0\)

\(\,\,\,\,\,\displaystyle\frac{2A}{h}-b=k \,\,\, h\ne 0\)

\(\,\,\,\,\,k = \displaystyle\frac{2A}{h}-b \,\,\, h\ne 0\)

\(\textbf{2)}\) \(\text{Solve for }C. \,\, F=\frac{9}{5}C+32\)

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The answer is \(C=\frac{5}{9}(F-32)\)

\(\textbf{3)}\) \(\text{Solve for }b. \,\, A=\frac{1}{2}bh\)

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The answer is \(b=\frac{2A}{h}, \,\,\, h\ne 0\)

\(\textbf{4)}\) \(\text{Solve for }x. \,\, y=3x+5\)

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The answer is \(x=\frac{y-5}{3}\)

\(\textbf{5)}\) \(\text{Solve for }m. \,\, R=2m-5\)

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The answer is \(m=\frac{R+5}{2}\)

\(\textbf{6)}\) \(\text{Solve for }d. \,\, G=2d+3c\)

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The answer is \(d=\frac{G-3c}{2}\)

\(\textbf{7)}\) \(\text{Solve for }s. \,\, F=\frac{1}{5}s-5\)

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The answer is \(s=5F+25\)

\(\textbf{8)}\) \(\text{Solve for }j. \,\, w=3j+2h-32\)

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The answer is \(j=\frac{w-2h+32}{3}\)

\(\textbf{9)}\) \(\text{Volume of a sphere: } V=\frac{4}{3} \pi r^3 \text{, solve for }r \)

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The Answer is \( r=\sqrt[3]{\frac{3V}{4 \pi}} \)

\(\textbf{10)}\) \(\text{Area of a trapezoid: } A=\frac{1}{2} \left(b_1+b_2\right)h \text{, solve for }h \)

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The Answer is \( h=\frac{2A}{b_1+b_2} \)

\(\textbf{11)}\) \(\text{Area of a trapezoid: } A=\frac{1}{2} \left(b_1+b_2\right)h \text{, solve for }b_1 \)

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The Answer is \( b_1=\frac{2A}{h}-b_2 \)

\(\textbf{12)}\) \(\text{Area of a circle: } A=\pi r^2 \text{, solve for }r \)

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The Answer is \( r=\sqrt{\frac{A}{\pi}} \)

\(\textbf{13)}\) \(\frac{a+b}{c+b}=2, \text{ solve for }b \)

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The Answer is \( b=a-2c\)

\(\textbf{14)}\) \(m^2+bx^2=k, \text{ solve for }x \)

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The Answer is \( x=\pm \sqrt{\frac{k-m^2}{b}}\)

Challenge Problems

\(\textbf{15)}\) \(\text{Solve for }y. \,\, hy+by=x\)

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The answer is \(y=\frac{x}{h+b}, \,\,\, h\ne -b\)

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\(\,\,\,\,\,hy+by=x\,\,\,\left(\text{Given}\right)\)

\(\,\,\,\,\,y(h+b)=x\,\,\,\left(\text{Factor out a y}\right)\)

\(\,\,\,\,\,y=\frac{x}{h+b}\,\,\,\left(\text{divide both sides by (h+b)}\right)\)

The answer is \(y=\frac{x}{h+b}, \,\,\, h\ne -b\)

\(\textbf{16)}\) If \(\frac{x}{y}=3\) and \(xz+1=16\), then \(yz=\)

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The answer is \(yz=5\)

\(\textbf{17)}\) If \(y=xz+x\). What is \( \displaystyle\frac{y}{x}\) ?

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The answer is \(\frac{y}{x}=z+1, \,\,\, x\ne 0\)

\(\textbf{18)}\) If \(\frac{2}{x}=8\), then \(3x=\)

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The answer is \(3x=\displaystyle\frac{3}{4}\)

\(\textbf{19)}\) If \(4p+3q=7b\), what is \(q\) in terms of \(b\) and \(p\)?

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The answer is \(q=\displaystyle\frac{7b-4p}{3}\)

\(\textbf{20)}\) Solve for y,
\(zy+2=3y+x\)

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The answer is \(y=\displaystyle\frac{x-2}{z-3}, \,\,\, z\ne 3\)

See Related Pages\(\)

\(\bullet\text{ Fahrenheit and Celsius Conversions}\)
\(\,\,\,\,\,\,\,\,F=\frac{9}{5}C+32…\)

In Summary

To isolate a variable in an algebraic equation means to rewrite the equation in a form where the variable appears on one side of the equal sign and all the other terms appear on the other side. This is useful for solving equations and understanding the behavior of the variable.

To isolate a variable in an equation, you can use the basic operations of algebra, such as addition, subtraction, multiplication, and division. The specific steps you need to take will depend on the form of the equation and the variable you are trying to isolate.

Isolating a variable in an equation is an important concept in algebra, as it allows you to solve equations and understand the behavior of the variable.

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