Question

How do you solve by substitution $a=3b-4\text{and}a+b=16?$

Answer 1

Hint: In substitution method, we take the value of a variable from an equation in terms of other variable and then put the value of that variable in another equation and after solving that another equation, we will get the value of the other variable, so we put its value in any of the equations to get the value of the first variable.

Complete step by step solution:
In order to solve the given equations $a=3b-4\text{and}a+b=16$ through substitution method, we will proceed as follows:
From one of the two equations, we will take value of one variable in terms of other,
$\Rightarrow a=3b-4-----(i)$
Now putting $a=3b-4$ in place of a in the second equation, we will get
$\Rightarrow (3b-4)+b=16$
We will now solve this equation for value of b
$$\begin{gathered} \Rightarrow 3b-4+b=16 \\ \Rightarrow 4b-4=16 \end{gathered}$$
Adding four to both sides of the equation, we will get
$$\begin{gathered} \Rightarrow 4b-4+4=16+4 \\ \Rightarrow 4b=20 \end{gathered}$$
Now, dividing both sides of the equation with coefficient of b, we will get
$$\begin{gathered} \Rightarrow {\displaystyle \frac{4b}{4}}={\displaystyle \frac{20}{4}} \\ \Rightarrow b=5 \end{gathered}$$
So we got the value of $b=5$
Now putting the value of b in equation (i), to get the value of the other variable, that is a
$$\begin{gathered} \Rightarrow a=3\times 5-4 \\ \Rightarrow a=15-4 \\ \Rightarrow a=11 \end{gathered}$$
Therefore $a=11\text{and}b=5$ is the required solution of the given equations.

Note: When using substitution method, many students substitute the value of the variable in the equation from where they have taken the value of that variable, which gives them the equation $0=0$ which is no wonder. Because they are putting the value in the parent equation from where they have taken the value. Never substitute the value in the equation from where it has taken.