Question

How do you solve $5x+y=5$ and $3x+2y=3$ using substitution?

Answer 1

Hint: In this question we have two linear equations which have the same solutions. These types of equations are called simultaneous equation and we will use method of substitution to solve for the values of $x$ and $y$ . we will first rearrange the first equation to get the value of $y$in terms of $x$ and then we will substitute it in the next equation to solve for the value of $x$, then we will put the value of $x$ found again in the first equation to get the value of $y$.

Complete step by step answer:
We have the equations:
$5x+y=5\to (1)$
$3x+2y=3\to (2)$
Now consider equation $(1)$
$\Rightarrow 5x+y=5$
We will solve for the value of $y$.
On transferring the term $5x$ from the left-hand side to the right-hand side, we get:
$\Rightarrow y=5-5x\to (3)$
Now consider equation $(2)$
$\Rightarrow 3x+2y=3$
Now we will substitute the value of $y=5-5x$ in equation $(2)$ and solve for $x$ as:
$\Rightarrow 3x+2(5-5x)=3$
On simplifying the brackets, we get:
$\Rightarrow 3x+10-10x=3$
Now on transferring the term $10$ from the left-hand to the right-hand side, we get:
$\Rightarrow 3x-10x=3-10$
On simplifying the terms, we get:
$\Rightarrow -7x=-7$
On multiplying both sides of the equation with $-1$ , we get:
$\Rightarrow 7x=7$
On transferring $7$from the left-hand side to the right-hand side, we get:
$\Rightarrow x=\dfrac{7}{7}$
On simplifying, we get:
$\Rightarrow x=1$ , which is the required value of $x$ .
On substituting $x=1$ in equation $(3)$ , we get:
$\Rightarrow y=5-5(1)$
On simplifying, we get:
$\Rightarrow y=0$, which is the required value of $y$.

Therefore, the value of $x=1$ and $y=0$ which is the required solution.

Note: To check whether the solution is correct we have to test the values of $x$ and $y$ in equation $(1)$ and $(2)$
On substituting the values in the left-hand side of equation $(1)$, we get:
$\Rightarrow 5(1)+0$
On simplifying we get:
$\Rightarrow 5$
Which is equal to the right-hand side, therefore the solution is correct for equation $(1)$ .
On substituting the values in the left-hand side of equation $(2)$, we get:
$\Rightarrow 3(1)+2(0)$
On simplifying we get:
$\Rightarrow 3$
Which is equal to the right-hand side, therefore the solution is correct for equation $(2)$ also.
In the given question we had two variables which are $x$ and $y$ , therefore they can be solved by using elimination or substitution, where there are more than three variables, the matrix is used to solve them.