Question

How do you solve the system of equations ${x^2} + {y^2} = 34$ and $y = x + 2$ algebraically?

Answer 1

We will first put in $y = x + 2$ in the equation ${x^2} + {y^2} = 34$. Then, we will obtain a quadratic equation in $x$. Then, we will use the quadratics formula to find the possible values of x and thus y.

Complete step by step solution:
We are given that we are required to solve the system of equations ${x^2} + {y^2} = 34$ and $y = x + 2$ algebraically.
Let us term the equation ${x^2} + {y^2} = 34$ as equation number 1 and equation $y = x + 2$ as equation number 2.
Now, we will just put in the value of y from equation number 2 in equation number 1, we will then obtain the following expression with us:-
$\Rightarrow {x^2} + {(x + 2)^2} = 34$ ………(3)
We know that we have a formula given by the following expression:-
$\Rightarrow {(a + b)^2} = {a^2} + {b^2} + 2ab$
Replacing a by x and b by 2 in the above mentioned formula, we will then obtain the following expression:-
$\Rightarrow {(x + 2)^2} = {x^2} + {2^2} + 2 \times 2 \times x$
Simplifying the calculations in the right hand side of the above expression, we will then obtain the following expression:-
$\Rightarrow {(x + 2)^2} = {x^2} + 4 + 4x$
Putting the above derived value in equation number 3, we will then obtain the following expression:-
$\Rightarrow {x^2} + {x^2} + 4 + 4x = 34$
Simplifying the left hand side of the above expression, we will then obtain the following expression:-
$\Rightarrow 2{x^2} + 4 + 4x = 34$
Taking 4 from addition in the left hand side to subtraction in the right hand side, we will then obtain the following expression:-
$\Rightarrow 2{x^2} + 4x = 34 - 4$
Simplifying the right hand side of the above expression, we will then obtain the following expression:-
$\Rightarrow 2{x^2} + 4x = 30$
We can write this expression as follows:-
$\Rightarrow {x^2} + 2x - 15 = 0$
Now, we know that if we have an equation given by $a{x^2} + bx + c = 0$ , then its roots are given by the following expression:-
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Putting a = 1, b = 2 and c = - 15, we will then obtain the following expression:-
$\Rightarrow x = \dfrac{{ - (2) \pm \sqrt {{{(2)}^2} - 4(1)( - 15)} }}{{2(1)}}$
Simplifying the squares and multiplications in the above expression, we will obtain:-
$\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 + 60} }}{2}$
Solving it, we will then obtain:-
$\Rightarrow x = 3, - 5$
Now, putting this in equation number 2, we will then obtain the following expression:-
$\Rightarrow y = 5, - 3$

Hence, the required answer is (3, 5) and (-5, -3).

Note:
The students must note that there is an alternate way to solve the same question.
Alternate way:
Here, we will use the method of splitting of the middle term as given below:
$\Rightarrow 2{x^2} + 4x - 30 = 0$
Now, we see that 2, 4 and 30, all of these numbers have 2 common in them. Therefore, we will take out 2 common from all of these and obtain the following expression:-
$\Rightarrow {x^2} + 2x - 15 = 0$
Now, we will break the middle term and then, we can write it as the following expression:-
$\Rightarrow {x^2} - 3x + 5x - 15 = 0$
Now, we can take x common out of the first two terms in the bracket, then we will obtain the following expression with us:-
$\Rightarrow x\left( {x - 3} \right) + 5x - 15 = 0$
As we did in the last step, we can also take out 5 common from the last two terms in the bracket above, we will then obtain the following expression with us:-
$\Rightarrow x\left( {x - 3} \right) + 5\left( {x - 3} \right) = 0$
Now, since we have (x – 3) in both the terms in the above expression, we can take it out as well to obtain the following expression with us:-
$\Rightarrow \left( {x - 3} \right)\left( {x + 5} \right) = 0$
Thus, we have the required answer as we did in the solution.