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Equations in several variables in practice – The linear case

Solving exercises about systems of equations
Hello. Let us solve the first exercise of the “Equations in several variables in practice” step. OK, we have to solve this system. First of all, observe that this a linear system or a system of linear equations. What it means that this is a linear system means that in both the equations, each segment is a term of degree at most 1. OK, then first of all, it is linear. And because it is linear, we have immediately a good instrument to understand if this linear system admits solutions and if it admits 1 or infinite solutions. Which is the idea? The idea is to compute the determinant of the system. How do you compute the determinant of the system?
You have to take the coefficient of x in the first equation and to multiply it by the coefficient of y in the second equation. And then we subtract the coefficient of y in the first equation times the coefficient of x in the second equation. OK, you get 5 times 2, 10, minus 3, 7, which is different from 0. Therefore, we have a unique solution for our system. And now, to find the solutions of a linear system, we have a couple of methods. The substitution method and the elimination method. The idea for the substitution method is to choose one of these two equations. And from that equation, just to get one variable in function of the other variable.
And then to substitute what you get in the other equation. Or otherwise, there is the elimination method. In this case, we will try to follow this second method. Which is the idea? The idea is to modify the equations in such a way that for at least one variable both they have the same coefficient. You see, for example, here, x has co-efficient 5, and here x has coefficient 1. But if we multiply the second equation by 5, then, also in the second equation, x will have as coefficient 5. OK, let us try. Then I repeat the first equation, 5x plus 3 times y equal 2. And we multiply by 5 the second equation.
We get 5x plus 10 times y equal to minus 25. OK, and now what do we do? And now we just take the subtraction of these two equations. For example, we can take the second one minus the first one. And what do we get? OK, we get 5x minus 5x, which is 0. 10 y minus 3y, which is 7y. And minus 25 minus 2, which is minus 27. You see. What happened? Happened that one variable, the x, disappeared. And now we have a very simple equation in only one variable. And we start solving this equation. What do we get? y equal to minus 27 over 7, OK. And now– OK, now we know the value of the variable y.
And from one of these two equations, we can get the value of x. How? For example, take the first equation, and we substitute this value to y. And what do we get? We get 5 times x plus 3 times minus 27 over 7 equal 2. You see, now this is, again, an equation in only one variable. And therefore, we can easily solve it. OK, let us try. x, we have 5 times x equal to 2 minus this quantity. But this quantity is negative. Therefore, minus this quantity will be just the module of this. And therefore, we get plus 3 times 27. OK, 20 times 3 is 60. 7 times 3 is 21.
60 plus 21 is 81 plus 81 over 7. OK, which is– we can take the common denominator 7 and we get 14 plus 81, which is 81 plus 14 is 95 over 7. OK, and now, it is sufficient to divide by 5 on both sides to get x equal 2. OK, 5, 95 divided 5. If you take 100 divide 5 is 20. Therefore, 95 divided by 5 will be 19. Therefore, we get x equal to 19 over 7. Then the solution of our system is the – attention – the pair, 19 over 7, and minus 27 over 7. This is the solution. This is– the first one is the value for the variable x.
The second one is the value for the variable y. Thank you, for your attention.

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 1.

Solve the system (quadleftlbrace begin{array}{l} 5x+3y = 2 x+2y= -5 end{array} right.)

Exercise. 2 [Solved only in the PDF file]

Solve the following systems:

i) (quadleftlbrace begin{array}{l} x-2y = 3 4x-8y = 12 end{array} right.)

ii) (quadleftlbrace begin{array}{l} x-2y = 1 -2x+4y = 5 end{array} right.)

Ex 3. [Solved only in the PDF file]

Discuss the value of the parameter (hinmathbb{R}) in such a way that the following systems of polynomial equations of degree 1 in the unknowns (x), (y) admit one solution, infinite solutions, no solutions.

i) (quadleftlbrace begin{array}{l} hx+y = 1 x-y = 2 end{array} right.)

ii) (quadleftlbrace begin{array}{l} x+y = 6 x+hy = 2h end{array} right.)

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Precalculus: the Mathematics of Numbers, Functions and Equations

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