Question

Solve the system of equations shown below algebraically?
${(x - 3)^2} + {(y + 2)^2} = 16$
$2x + 2y = 10$

Answer 1

Hint: In this question, we will simplify the equation. We will convert equation 2 in y firm and substitute the value of y in equation 1. Then we will get a quadratic equation. The general form of the quadratic equation is$a{x^2} + bx + c = 0$. Where ‘a’ is the coefficient of${x^2}$, ‘b’ is the coefficient of x and ‘c’ is the constant term.
To solve this equation, we will apply the sum-product pattern. During the simplification, we will take out common factors from the two pairs. Then we will rewrite it in factored form.
Therefore, we should follow the below steps:
Apply sum-product patterns.
Make two pairs.
Common factor from two pairs.
Rewrite in factored form.

Complete step by step solution:
In this question, two equations are given.
${(x - 3)^2} + {(y + 2)^2} = 16$ ...(1)
$2x + 2y = 10$ ...(2)
Take equation (2) and simplify it.
$ \Rightarrow 2x + 2y = 10$
Take out 2 as a common factor from the left-hand side.
$ \Rightarrow 2\left( {x + y} \right) = 10$
Divide both sides by 2.
$ \Rightarrow \dfrac{{2\left( {x + y} \right)}}{2} = \dfrac{{10}}{2}$
That is equal to,
$ \Rightarrow x + y = 5$
Let us subtract x on both sides.
$ \Rightarrow x - x + y = 5 - x$
That is equal to,
$ \Rightarrow y = 5 - x$
Now, put the value of y in equation (1).
$ \Rightarrow {(x - 3)^2} + {(y + 2)^2} = 16$
Put $y = 5 - x$.
 $ \Rightarrow {(x - 3)^2} + {(5 - x + 2)^2} = 16$
Therefore,
$ \Rightarrow {(x - 3)^2} + {(7 - x)^2} = 16$
Apply the algebraic identity ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
Therefore,
$ \Rightarrow {x^2} - 2\left( x \right)\left( 3 \right) + {\left( 3 \right)^2} + {(7)^2} - 2\left( x \right)\left( 7 \right) + {x^2} = 16$
Let us simplify the above expression.
$ \Rightarrow 2{x^2} - 6x + 9 + 49 - 14x - 16 = 0$
So,
$ \Rightarrow 2{x^2} - 20x + 42 = 0$
Take out 2 as a common factor from the left-hand side.
$ \Rightarrow 2\left( {{x^2} - 10x + 21} \right) = 0$
Therefore,
$ \Rightarrow {x^2} - 10x + 21 = 0$
Here, this is the quadratic equation.
Let us apply the sum-product pattern in the above equation.
Since the coefficient of ${x^2}$is 1 and the constant term is 21. Let us multiply 1 and 21. The answer will be 21. We have to find the factors of 21 which sum to -10. Here, the factors are -3 and -7.
Therefore,
$ \Rightarrow {x^2} - 3x - 7x + 21 = 0$
Now, make two pairs in the above equation.
$ \Rightarrow \left( {{x^2} - 3x} \right) - \left( {7x - 21} \right) = 0$
Let us take out the common factor.
$ \Rightarrow x\left( {x - 3} \right) - 7\left( {x - 3} \right) = 0$
Now, rewrite the above equation in factored form.
$ \Rightarrow \left( {x - 7} \right)\left( {x - 3} \right) = 0$
Now,
$ \Rightarrow \left( {x - 7} \right) = 0$ and $ \Rightarrow \left( {x - 3} \right) = 0$
Simplify them.
$ \Rightarrow x - 7 + 7 = 0 + 7$ and $ \Rightarrow x + 3 - 3 = 0 + 3$
That is equal to,
$ \Rightarrow x = 7$ and $ \Rightarrow x = 3$

Hence, the roots of the given equation are 7 and 3.

Note:
One important thing is, we can always check our work by multiplying out factors back together, and check that we have got back the original answer.
$ \Rightarrow x = 7$ and $ \Rightarrow x = 3$
Simplify them.
$ \Rightarrow x - 7 = 7 - 7$ and $ \Rightarrow x - 3 = 3 - 3$
 That is equal to,
$ \Rightarrow \left( {x - 7} \right) = 0$ and $ \Rightarrow \left( {x - 3} \right) = 0$
To check our factorization, multiplication goes like this:
$ \Rightarrow \left( {x - 7} \right)\left( {x - 3} \right) = 0$
Let us apply multiplication to remove brackets.
$ \Rightarrow {x^2} - 3x - 7x + 21 = 0$
Let us simplify it. We will get,
$ \Rightarrow {x^2} - 10x + 21 = 0$
Hence, we get our quadratic equation back by applying multiplication.