Question

How do you solve the simultaneous equations \[{x^2} + {y^2} = 29\] and \[y - x = 3\] ?

Answer 1

Hint: In this question you have to find the value of x and y that is \[\left( {x,y} \right)\] of the given two equations by the method of substitution of simultaneous equations. We can’t solve this by the elimination method so we solve by the substitution method. On further simplification we can obtain the required solution.

Complete step-by-step answer:
Here in this question we have two equations. One equation is having the degree 2 and the other one is having the degree 1. So we can’t solve this question by the elimination method. So to solve this equation we use a substitution method. So let we solve the equations as simultaneous equations using the method of substitution
consider the given equations
 \[{x^2} + {y^2} = 29\] -------(1)
 \[y - x = 3\] --------(2)
Take the -x to the RHS, then the equation (2) is written as
 \[y = x + 3\] ----- (3)
Substitute the equation (3) in equation (1) , then we have
 \[ \Rightarrow {x^2} + {(x + 3)^2} = 29\]
Let we use the standard algebraic formula \[{(a + b)^2} = {a^2} + 2ab + {b^2}\] , by using this the above equation is written as
 \[ \Rightarrow {x^2} + {x^2} + 6x + 9 = 29\]
On simplifying we have
 \[ \Rightarrow 2{x^2} + 6x - 20 = 0\]
Divide the above equation by 2 we have
 \[ \Rightarrow {x^2} + 3x - 10 = 0\]
To find the value of x we use the factorisation method.
So the above equation is written as
  \[ \Rightarrow {x^2} + 5x - 2x - 10 = 0\]
Let we take x as common in first two terms and -2 as common in the last two terms and the equation is rewritten as
 \[ \Rightarrow x(x + 5) - 2(x + 5) = 0\]
Take (x+5) as common in the above equation we have
 \[ \Rightarrow (x + 5)(x - 2) = 0\]
On simplification we have
 \[ \Rightarrow x + 5 = 0\] and \[x - 2 = 0\]
Therefore,
 \[ \Rightarrow x = - 5\] and \[x = 2\]
When x is -5, Substituting in the equation (2) we have
 \[
  y - ( - 5) = 3 \\
   \Rightarrow y + 5 = 3 \\
   \Rightarrow y = 3 - 5 \\
   \Rightarrow y = - 2 \;
 \]
Suppose when the value of x is 2, substituting in the equation (2) we have
 \[
  y - 2 = 3 \\
   \Rightarrow y = 3 + 2 \\
   \Rightarrow y = 5 \;
 \]
Hence, we have simultaneous equation and found the value of x and y. so we have (-5, -2) and (2, 5)
So, the correct answer is “(-5, -2) and (2, 5) ”.

Note: to solve the simultaneous equation we go through the elimination method. We follow the elimination method when the both equations are having the same degree. But here the two equations have different degrees. So it is highly impossible to apply the elimination method.