Question

How do you solve the simultaneous equations $7a - 3b = 17$ and $2a + b = 16$?

Answer 1

In this question, we are given with two equations and are asked to solve it simultaneously. The two equations are $7a - 3b = 17$ and $2a + b = 16$.
Since there are two different variables in both the equations, we can substitute one in another and find the values of the variables. On doing some simplification we get the required answer.

Complete step by step answer:
We are given with two equations and are asked to solve it simultaneously.
The two given equations are $7a - 3b = 17$ and $2a + b = 16$.
First we will find an equation for any one variable and then substitute it in the other equation to find the value of the variables.
Now we got the two different equations which are: $7a - 3b = 17$ and $2a + b = 16$
Now we will use substitution method:
From$2a + b = 16$ we can say that $b = 16 - 2a$ by transferring the other variable to the right hand side.
Now as we know $b = 16 - 2a$, we can substitute this in the first equation
Substituting $b = 16 - 2a$ in $7a - 3b = 17$we get,
$\Rightarrow 7a - 3b = 17$
$\Rightarrow 7a - 3(16 - 2a) = 17$
Multiplying the brackets and the number,
$\Rightarrow 7a - 48 + 6a = 17$
Rearranging the equation
$\Rightarrow 7a + 6a - 48 = 17$
Adding the variables,
$\Rightarrow 13a - 48 = 17$
Now, transferring $48$ to the other side,
$\Rightarrow 13a = 17 + 48$
$\Rightarrow 13a = 65$
Dividing $13$ on both sides we get,
$\Rightarrow a = \dfrac{{65}}{{13}}$
Therefore the value of $a$ is,
$a = 5$
As we got the value of $a$ , we will now substitute this in $b = 16 - 2a$ and find the value of $b$.
Put $a = 5$
$\Rightarrow b = 16 - 2a$
$\Rightarrow b = 16 - 2(5)$
Multiplying the bracket,
$\Rightarrow b = 16 - 10$
$\Rightarrow b = 6$

Therefore, the values of $a$ and $b$ are $5$ and $6$ .

Note: Alternative method:
We can find the values of $a$ and $b$ by another method also,
As we know the two equations, $7a - 3b = 17$ and $2a + b = 16$
We can either add or subtract it in order to get the value of any one of the variables.
We have to do it in such a way that one variable should cancel out itself. We can alter or rewrite the equations for that purpose.
Here we can see that the coefficients of the variables in the equations are different. In order to make it cancel we need to rewrite the equation in such a way that both the coefficient of the equations should be equal.
Changing the second equation by multiplying it by $3$, we get
$6a + 3b = 48$
Now adding the both equation, we get
${{\text{7}}a - 3b = 17} \\\ {\underline {{\text{6}}a + 3b = 16{\text{( + )}}} } \\\ {{\text{13}}a = 48}$
As we can see there are two positive $a$ so it becomes $13a$ when added. And there is one $+ b$ and one $- b$, so when they are added, they cancel out and become$0$ . In this way we can find the value of other variables.
So now, finding $a$,
$a = \dfrac{{48}}{{13}}$
We get, $a = 5$
Now substituting the value of $a$ in any of the equations, we can find $b$ .
Let’s put $a = 5$in the second equation.
$\Rightarrow 2a + b = 16$
$\Rightarrow b = 16 - 10$
Let us subtract we get
$\Rightarrow b = 6$
Therefore the values of $a$ and $b$ are $5$ and $6$ .