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+ = , (1)
x/2 + y/5 = 5. (2)
This system of equations in non-linear.
To make the solution easier, I will introduce NEW VARIABLES a = , b = . Then the system takes the form
a + b = (3)
+ = 5 (4)
To solve it, first simplify equation (4):
+ = 5 ====> = 5 ====> replace a+b by , based on (3) ====> = 5 ====> ab = .
Next, from (3), express b = - a and substitute it into equation ab = . You will get
= ,
-6a^2 + 5a = 1
6a^2 - 5a + 1 = 0.
Solve the last quadratic equation using the quadratic formula.
You will get two solutions: a= and a= .
Thus, the system (3)-(4) has two solutions:
1) a= , b= - = = ,
and
2) a = , b= - = = = .
Now you need to return from "a" and "b" to x and y, via the formulas a = , b = .
By doing so, you get two solutions for the original system:
1) x = = = 4; y = = = 15.
2) x = = = 6; y = = = 10.
Answer. The system has two solutions 1) x= 4; y= 15 and 2) x= 6, y= 10.
Solved.
You may check that the solution is correct by substituting the found values into the original equations.
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I am very glad that the tutor @MathLover1 placed her solution here.
Comparing these two, you can see how many tons of calculations I saved you from, using my substitutions ! ! !
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To see other similar solved problems for systems of two non-linear equations in two unknowns, look into the lessons
- Solving systems of non-linear equations by reducing to linear ones
- Solving systems of non-linear equations in two unknowns using the Cramer's rule
in this site.