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Equations in several variables in practice – The NON linear case

Solving exercises about systems of equations
Hello. Let us consider the first exercise of the “equations in several variables in practice” step. We have to solve, now, this system. You immediately realize that this is not a linear system. Indeed, in the second equation, it’s not true that any segment is a term of degree at most 1. For example, this term has degree 3.
When you have to solve a system like this one, probably the more reasonable method is the substitution method. Let us see how to solve this system. You see, we have two equations. One is simpler than the other. The first one is simpler, of course. And through the first one, you easily can write y in function of x. Indeed, you have that y is equal to 4 minus x. Then, now, we can substitute in the second equation, in a more complicated one. We can substitute to y 4 minus x. And you get x cubed plus 4 minus x cubed equal 28. And you see what happens, as always with the substitution method, we get one equation with only one variable.
Let us try to find the solutions of this equation. Good.
Then we have x cubed plus. And we have to take the cube of this binomial. And that is 4 to the cube. Then we have minus 3 times the square of 4 times x, then we have plus 3 times 4 times the square of minus x, or of x, which is the same. And finally, minus x cubed, and this is to be equal to 28. Now, you immediately realize that this term and this term can be erased, and we remain with an equation of degree 2 in x. And the equation is the following. You have 12 times x square, this term. Then we have this one. The square of 4 is 16 times 3, and you get minus 48.
And then we have the constant term, which is the cube of 4. Then you have to do 4 times 4, which is 16, times 4. 16 times 4 is 64.
And then you have this term, which becomes minus 28, equals 0. 64 minus 28 is equal to 36.
You immediately realise that all these coefficients are divisible by 12. Therefore, we can rewrite this equation in a simpler way. Dividing by 12, you have x square minus 4 times x plus 3 equals 0. OK now we have an equation of degree 2, and we compute its discriminant, which is the square of minus 4 minus 4 times the leading coefficient, which is 1, times the constant term.
Of course, it is 16 minus 12, which is 4. The discriminant is greater than 0. Therefore, this equation will have two solutions. Which are the solutions? [They] are x equal to minus minus 4, which is 4, plus or minus the square root of the discriminant over 2 times the leading coefficient. Then we get 4 plus 2 divided by 2, which is 3, or 4 minus 2 divided by 2, which is 1. These are the two possible solutions for our equation of the second degree. Therefore, now, we can obtain the values of y, remembering that y is equal to 4 minus x.
Then if x is equal to 3, then y will be 4 minus 3, will be equal to 1. If x is equal to 1, then y will be 4 minus 1, 3. Therefore, the solutions of this system are the following pairs– the pair 3, 1, and the pair 1, 3.
Ciao a tutti.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video.

In any case, you will find below a PDF file with the solutions.

Exercise 4.

Solve the system (quadleftlbrace begin{array}{l} x+y=4 x^3+y^3=28 end{array} right. )

Exercise. 5 [Solved only in the PDF file]

Solve the following systems:

i) (quadleftlbrace begin{array}{l} x^2+y^2=13 xy=-6 end{array} right. )

ii) (quadleftlbrace begin{array}{l} sqrt{2-x}=y y+2=x end{array} right. )

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Precalculus: the Mathematics of Numbers, Functions and Equations

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