Question

How do you solve \[y + 2x = 5\] and \[2{x^2} - 3x - y = 16\] ?

Answer 1

Hint: We have two equations. Now we use the equation \[y + 2x = 5\] we find the value of ‘y’ in terms of ‘x’ and constant. We substitute it in the equation \[2{x^2} - 3x - y = 16\] . After simplification we will have a quadratic equation and we solve it using factorization method or by quadratic formula.

Complete step by step solution:
Given,
 \[y + 2x = 5 - - \left( 1 \right)\]
 \[2{x^2} - 3x - y = 16 - - - \left( 2 \right)\]
Now from (1) we have,
 \[y = 5 - 2x\]
Substituting this in equation (2) we have,
 \[2{x^2} - 3x - y = 16\]
 \[2{x^2} - 3x - \left( {5 - 2x} \right) = 16\]
 \[2{x^2} - 3x - 5 + 2x = 16\]
 \[2{x^2} - 3x - 5 + 2x - 16 = 0\]
 \[ \Rightarrow 2{x^2} - x - 21 = 0\]
Thus we have quadratic equation and when comparing with \[a{x^2} + bx + c = 0\] we have, \[a = 2,b = - 1,c = - 21\] ,
Now substituting this in \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] , we have,
 \[x = \dfrac{{ - \left( { - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( 2 \right)\left( { - 21} \right)} }}{{2\left( 2 \right)}}\]
 \[x = \dfrac{{1 \pm \sqrt {1 + 168} }}{4}\]
 \[x = \dfrac{{1 \pm \sqrt {169} }}{4}\]
 \[x = \dfrac{{1 \pm 13}}{4}\]
Thus we have two roots,
 \[ \Rightarrow x = \dfrac{{1 + 13}}{4}\] and \[x = \dfrac{{1 - 13}}{4}\]
 \[ \Rightarrow x = \dfrac{{14}}{4}\] and \[x = \dfrac{{ - 12}}{4}\]
 \[ \Rightarrow x = \dfrac{7}{2}\] and \[x = - 3\]
Now to find the ‘y’ value we substitute \[ \Rightarrow x = \dfrac{7}{2}\] and \[x = - 3\] in any one of the above two equations.
Let’s substitute in equation (1),
Put \[x = \dfrac{7}{2}\] in \[y + 2x = 5\]
 \[y + 2\left( {\dfrac{7}{2}} \right) = 5\]
 \[y + 7 = 5\]
 \[y = 5 - 7\]
 \[ \Rightarrow y = - 2\]
Now put \[x = - 3\] in \[y + 2x = 5\] ,
 \[y + 2\left( { - 3} \right) = 5\]
 \[y - 6 = 5\]
 \[y = 5 + 6\]
 \[ \Rightarrow y = 11\] .
Thus the solution of \[y + 2x = 5\] and \[2{x^2} - 3x - y = 16\] are \[ x = \dfrac{7}{2}, y = - 2\] and \[ x = - 3, y = 11\] .
So, the correct answer is \[ x = \dfrac{7}{2}, y = - 2\] and \[ x = - 3, y = 11\] .

Note: We know that if we have a polynomial of degree n, then that polynomial will have n roots or n factors. We call roots as zeros also. We use the quadratic formula when the factorization method files. In above the factorization method fails hence we have used that formula.