Question

How do you solve $ - x + y = 1$ and ${x^2} + {y^2} = 16$ using substitution?

Answer 1

Hint: In order to solve the Equations $ - x + y = 1$ and ${x^2} + {y^2} = 16$ using substitution take first equation then write $y$ in terms of $x$. Put the value of $y$ in the Second equation, solve the Equation obtained and will get the value of $x$. Then put the obtained value of $x$ in $y$ and will get the value of y.

Formula used:
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
Determinant of quadratic equation $D = {b^2} - 4ac$

Complete step by step solution:
We are given with the Equations $ - x + y = 1$and ${x^2} + {y^2} = 16$.
In order to solve these Equations using Substitution method take the first equation and write $y$ in terms of $x$ and we get:
$y = 1 + x$.
Put this value of $y$ in the Second equation, and we get:
${x^2} + {\left( {1 + x} \right)^2} = 16$
Expand the Equations and solve:
$
  {x^2} + {\left( {1 + x} \right)^2} = 16 \\
\Rightarrow {x^2} + 1 + {x^2} + 2x = 16 \\
\Rightarrow 2{x^2} + 2x + 1 - 16 = 0 \\
\Rightarrow 2{x^2} + 2x - 15 = 0 \\
 $
We got a Quadratic Equation, solve it for $x$
Let’s first compare the given equation $2{x^2} + 2x - 15 = 0$with the standard quadratic equation $a{x^2} + bx + c$to get the values of $a,b,c$, we get
$
  a = 2 \\
  b = 2 \\
  c = - 15 \\
 $
Determinant $D$ of quadratic equation is given as $D = {b^2} - 4ac$
Putting the values of $a,b,c$, we get the determinant as
\[
  D = {\left( 2 \right)^2} - 4\left( 2 \right)\left( { - 15} \right) \\
 \Rightarrow D = 4 + 120 = 124 \\
 \]
Since, we got $D > 0$, which means there are two distinct real roots or in other words the equation has two x-intercepts.
The Root or intercepts of x-axis are $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
$
 {x_1} = \dfrac{{ - 2 + \sqrt {124} }}{{2\left( 2 \right)}} \\
\Rightarrow {x_1} = \dfrac{{ - 2 + 2\sqrt {31} }}{2} \\
\Rightarrow {x_1} = - 1 + \sqrt {31} \\
   {x_2} = \dfrac{{ - 2 - \sqrt {124} }}{{2\left( 2 \right)}} \\
\Rightarrow {x_2} = \dfrac{{ - 2 - 2\sqrt {31} }}{2} \\
\Rightarrow {x_2} = - 1 - \sqrt {31} \\
 $
Since, we obtained the value of $x$, put them in $y$, and get the value of $y$.
For 1st root of $x$, the value of $y$ is:
${y_1} = 1 + {x_1} = 1 + \left( { - 1 + \sqrt {31} } \right) = \sqrt {31} $
For 2nd root of $x$, the value of $y$ is:
${y_2} = 1 + {x_2} = 1 + \left( { - 1 - \sqrt {31} } \right) = - \sqrt {31} $
Therefore, after solving the two Equations $ - x + y = 1$ and ${x^2} + {y^2} = 16$ using substitution, we get
$\left( {{x_1},{y_1}} \right) = \left( { - 1 + \sqrt {31} ,\sqrt {31} } \right)$ and $\left( {{x_2},{y_2}} \right) = \left( { - 1 - \sqrt {31} , - \sqrt {31} } \right)$.

Note:
1) These Equations can be solved using Elimination or by graphing method instead of using Substitution method.
2) We can use mid-term factorization instead of Quadratic formula, we used This Quadratic formula because there was no possibility to use mid-term factorization in this equation.