Question 1057021
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<pre>
4x - y = 7,   (1)
xy = 15.      (2)

From (1), express y = 4x-7 and substitute it into equation (2), replacing y. You will get a single quadratic equation

x*(4x-7) = 15.

Simplify:

4x^2 - 7x - 15 = 0,

{{{x[1,2]}}} = {{{(7 +- sqrt(7^2 - 4*4*(-15)))/(2*4)}}} = {{{(7 +- sqrt(289))/8)}}} = {{{(7 +- 17)/8}}}.

{{{x[1]}}} = {{{24/8}}} = 3  --->  y = {{{15/x}}} = {{{15/3}}} = 5.

{{{x[2]}}} = {{{-10/8}}} = {{{-5/4}}} = -1.25.  --->  y = {{{15/x}}} = {{{15/((-5/4))}}} = -12.

<U>Answer</U>.  There are two solutions:  (x,y) = (3,5)  and  (x,y) = (-1.25,-12).
</pre>

For many other similar solved problems see the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-alg-eqns-of-deg2-deg1.lesson>Solving systems of algebraic equations of degree 2 and degree 1</A> 

in this site.


Also, you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lesson is the part of this online textbook under the topic "<U>Systems of equations that are not linear</U>".