Question

How do you solve for x and y in the simultaneous equation $ x+2y=3 $ and $ {{x}^{2}}-{{y}^{2}}=24 $

Answer 1

Hint: We will use the concept of substitution method to solve the above equation. At first, we will solve the first linear equation and find the value of x in terms of y i.e. $ x=3-2y $ and the we will substitute that x value in the second equation $ {{x}^{2}}-{{y}^{2}}=24 $ and then we will get: $ {{\left( 3-2y \right)}^{2}}-{{y}^{2}}=24 $ . Now, we will solve the equation and get the value of y and then we will substitute the value of y in the linear equation $ x+2y=3 $ to get the value of x.

Complete step by step answer:
We will use the method of substitution from algebra to solve the above simultaneous equation.
Since, we have to solve the equations $ x+2y=3 $ and $ {{x}^{2}}-{{y}^{2}}=24 $ to get the value of x and y.
At first, we will solve the given linear equation $ x+2y=3 $ to get the value of x in terms of y. We will take 2y to the RHS, then we will get:
 $ \Rightarrow x=2y-3---(1) $
Now, we will substitute the value of x from (1) in equation $ {{x}^{2}}-{{y}^{2}}=24 $ to convert the whole equation from two variable to one variable form.
 $ \Rightarrow {{\left( 3-2y \right)}^{2}}-{{y}^{2}}=24 $
Now, we will expand the quadratic equation and solve for y.
 $ \Rightarrow 9+4{{y}^{2}}-12y-{{y}^{2}}=24 $
 $ \Rightarrow 3{{y}^{2}}-12y-15=0 $
Now, we will divide the whole equation by 3 then we will get:
 $ \Rightarrow \dfrac{3{{y}^{2}}-12y-15}{3}=0 $
 $ \Rightarrow {{y}^{2}}-4y-5=0 $
Now, we will solve the above quadratic equation by using the middle term split method. We can split the middle term as -5y and y as (-5y + y) = -4y and $ -5\times 1=-5 $ which is equal to constant terms.
 $ \Rightarrow {{y}^{2}}-5y+y-5=0 $
Now, we will take common form ‘y’ the first two terms and 1 from the last two terms then we will get:
 $ \Rightarrow y\left( y-5 \right)+1\left( y-5 \right)=0 $
 $ \Rightarrow \left( y+1 \right)\left( y-5 \right)=0 $
 $ \Rightarrow y=5,-1 $
Therefore, the value of y is -1, 5.
Now, we will find the value of ‘x’ by substituting the value of ‘y’ in equation (1).
From equation (1) we have:
 $ x=3-2y $
So, when y = -1, we will get:
 $ \Rightarrow x=3-2\left( -1 \right) $
 $ \Rightarrow x=3+2=5 $
 $ \therefore x=5 $
Hence, we get x = 5, when y = -1.
Now, when y = 5, we will get:
 $ \Rightarrow x=3-2\left( 5 \right) $
 $ \Rightarrow x=3-10 $
 $ \therefore x=-7 $
Hence, when y = 5, we have x = -7.
Thus, value of x and y which satisfies the both equation $ x+2y=3 $ and $ {{x}^{2}}-{{y}^{2}}=24 $ is x = -7, and y = 5 and x = 5 and y = -1.
This is our required solution.

Note:
When we are given two simultaneous equations that contain two variables and the degree of both equations is not the same then we always use a substitution method to solve both equations to get the value of both variables. In the substitution method, we first express on of the variable in terms of another, and then we substitute it in other equation so that we get an equation in one variable and then we solve that equation to get the value of one variable, and then we will the value of one variable in any one equation given in the question to get the value of another variable.