Question

How do you solve by substitution \[x + y = 2\] and \[3x + 2y = 5\] ?

Answer 1

Hint: To solve this question we should know about linear equations. Linear equation in two variables: The equation having the highest degree on any variable. Be should keep in mind always LHS is equal to RHS if not then we are doing the wrong solution.

Complete step by step solution:
To solve this question we go step by step:
As given there are two linear equation:
 \[\Rightarrow x + y = 2\] ………… \[(1)\]
 \[\Rightarrow 3x + 2y = 5\] …………. \[(2)\]
We go step by step:
Step 1: We first take equation \[(1)\] and we try to find the value of $x$ with respect to $y$ .
So, \[\Rightarrow x + y = 2\]
Taking $x$ on one side:
 \[\Rightarrow x = 2 - y\]
Step 2: We put the calculated value of $x$ in \[(2)\] . We get,
 \[\Rightarrow 3x + 2y = 5\]
Keeping value in it,
 \[\Rightarrow 3\left( {2 - y} \right) + 2y = 5\]
Open the bracket,
 \[\Rightarrow 6 - 3y + 2y = 5\]
 \[ \Rightarrow 6 - y = 5\]
 \[ \Rightarrow 6 - 5 = y = 1\]
Step 3: keeping above calculated value in \[(1)\] . we get value of $x$ ,
 \[x + y = 2\]
Keeping \[y = 1\] ,
 \[ \Rightarrow x + 1 = 2\]
 \[ \Rightarrow x = 2 - 1 = 1\]

Hence, from the above calculation, $x = 1\,and\,y = 1$.

Note: Check the answer by putting values in given equation in the question:
 \[x + y = 2\]
Keep values in it,
 \[ \Rightarrow 1 + 1 = 2\]
 \[ \Rightarrow 2 = 2\]
Hence RHS=LHS so it is valid for the first Equation.
Check it for another equation.
 \[\Rightarrow 3x + 2y = 5\]
 \[ \Rightarrow 3.1 + 2.1 = 5\]
 \[ \Rightarrow 5 = 5\]
Hence RHS=LHS.
It is valid for both equations.
Hence our answer is true.