Algebra: Substitution Practice 2
Follow the three key steps diligently, and you’ll be able to solve substitution questions with full accuracy. Ready to put your skills to the test? Try these 5 practice questions to see how well you’ve master the method.
Related Lessons:
Practice 2 - Question 1
Given that \(a=2\), \(b=-3\) and \(c=-1\), evaluate \(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\).
Solution:
\[
\begin{align*}
\frac{a}{b}+\frac{c}{a}+\frac{b}{c}&=\frac{\left(2\right)}{\left(-3\right)}+\frac{\left(-1\right)}{\left(2\right)}+\frac{\left(-3\right)}{\left(-1\right)}\\
&=1\frac{5}{6}
\end{align*}
\]
Practice 2 - Question 2
Given that \(h=\frac{3u^2}{8g}\), calculate the value of g when \(u=32\) and \(h=7.9\).
Solution:
\[
\begin{align*}
h&=\frac{3u^2}{8g}\\
\left(7.9\right)&=\frac{3\left(32\right)^2}{8g}\\
&=\frac{3\left(32\right)^2}{8\times7.9}\\
&=48\frac{48}{79}
\end{align*}
\]
Practice 2 - Question 3
Given that \(p=2\), \(q=-3\) and \(r=-1\), evaluate \(\frac{r}{p+q}\).
Solution:
\[
\begin{align*}
\frac{r}{p+q}&=\frac{\left(-1\right)}{\left(2\right)+\left(-3\right)}\\
&=1
\end{align*}
\]
Practice 2 - Question 4
Given that \(x=4\), \(y=-4\) and \(z=2\), taking the positive square root, find the value of \(\frac{\sqrt x-y^2}{z}\).
Solution:
\[
\begin{align*}
\frac{\sqrt x-y^2}{z}&=\frac{\sqrt{\left(4\right)}-\left(-4\right)^2}{\left(2\right)}\\
&=-7
\end{align*}
\]
Practice 2 - Question 5
Given that \(x=-1.5\) and \(y=3.5\), find the value of \(\left(x-y\right)\).
Solution:
\[
\begin{align*}
\left(x-y\right)&=\left(\left(-1.5\right)-\left(3.5\right)\right)\\
&=-5
\end{align*}
\]
Step By Step Full Solution
Below is the video walkthrough covering the full step-by-step solutions for the five practice questions on algebraic substitution. The video clearly demonstrates the 3-step method used to evaluate values in an algebraic expression, guiding students through every stage of the substitution process.