# Solving linear equations Grade 7

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Sections: Maths

The author’s appeal to this topic is not accidental. Equations with two variables are first encountered in the 7th grade course. One equation with two variables has an infinite number of solutions. This graphically demonstrates the linear function graph defined in the form ax + by = c. In a school course, students learn systems of two equations with two variables. As a result, a whole series of problems drop out of the field of view of the teacher and, therefore, the student, with limited conditions on the coefficient of the equation, as well as methods for solving them.

It is about solving an equation with two unknowns in integers or natural numbers.

At school, natural and integer numbers are studied in grades 4-6. By the time they graduate, not all students remember the differences between the sets of these numbers.

However, a problem of the type “solve an equation of the form ax + by = c in integers” is increasingly encountered in entrance exams at universities and in exam papers.

The solution of indefinite equations develops logical thinking, ingenuity, attention to analyze.

I propose the development of several lessons on this topic. I have no definite recommendations on the timing of these lessons. Separate elements can be used in the 7th grade (for a strong class). These lessons can be taken as a basis and a small elective course on preprofile training in the 9th grade can be developed. And, of course, this material can be used in grades 10-11 to prepare for exams.

The purpose of the lesson:

repetition and generalization of knowledge on the topic “First and Second Order Equations”
• education of cognitive interest in the subject
• formation of skills to analyze, generalize, transfer knowledge to a new situation

### 4) Homework.

Examples. Solve the equation in integers:

but)

 2x = 4 2x = 5 2x = 5 x = 2 x = 5/2 x = 5/2 y = 0 not suitable not suitable 2x = -4 not suitable not suitable x = -2 y = 0

b)

at)

The results. What does it mean to solve the equation in integers?

What methods of solving indefinite equations do you know?

Exercises for training.

1) Decide in integers.

 a) 8x + 12y = 32 x = 1 + 3n, y = 2 - 2n, n Z b) 7x + 5y = 29 x = 2 + 5n, y = 3 - 7n, n Z c) 4x + 7y = 75 x = 3 + 7n, y = 9 - 4n, n Z d) 9x - 2y = 1 x = 1 - 2m, y = 4 + 9m, m Z d) 9x - 11y = 36 x = 4 + 11n, y = 9n, n Z f) 7x - 4y = 29 x = 3 + 4n, y = -2 + 7n, n Z g) 19x - 5y = 119 x = 1 + 5p, y = -20 + 19p, p Z h) 28x - 40y = 60 x = 45 + 10t, y = 30 + 7t, t Z

2) Find integer non-negative solutions to the equation:

 a) 8x + 65y = 81 x = 2, y = 1 b) 17x + 23y = 183 x = 4, y = 5

3) Find all pairs of integers (x, y) that satisfy the following conditions

 a) x + y = xy (0,0), (2,2) b) (1,2), (5,2), (-1,-1), (-5,-2)

The number 3 can be factorized:

 a) b) at) d)

 at) (11,12), (-11,-12), (-11,12), (11,-12) d) (24,23), (24,-23), (-24,-23), (-24,23) e) (48,0), (24,1), (24,-1) e) x = 3m, y = 2m, mZ g) y = 2x - 1 x = m: y = 2m - 1, m Z h) x = 2m, y = m, x = 2m, y = -m, m Z and) no solutions

4) Solve equations in integers

 (-3,-2), (-1,1), (0,4), (2,-2), (3,1), (5,4) (x - 3) (xy + 5) = 5 (-2,3), (2,-5), (4,0) (y + 1) (xy - 1) = 3 (0,-4), (1,-2), (1,2) (-4,-1), (-2,1), (2,-1), (4,1) (-11,-12), (-11,12), (11,-12), (11,12) (-24,23), (-24,23), (24,-23), (24,23)

5) Solve equations in integers.

 but) (-1,0) b) (5,0) at) (2,-1) d) (2, -1)

• Children's Encyclopedia “Pedagogy”, Moscow, 1972
• Algebra-8, N.Ya. Vilenkin, VO Nauka, Novosibirsk, 1992
• Competitive problems based on number theory. V.Ya. Galkin, D.Yu. Sychugov. Moscow State University, VMK, Moscow, 2005
• Problems of increased difficulty in the course of algebra 7-9 classes. N.P. Kosrykina. "Education", Moscow, 1991
• Algebra 7, Makarychev Yu.N., “Enlightenment”.
• ### How to solve the equation if “x” is negative

Often in the equations there is a situation when at "x" there is a negative coefficient. As, for example, in the equation below.

To solve such an equation, we ask ourselves the question again: “What do you need to divide“ −2 ”to get“ 1 ”?”. It must be divided by “−2”.

−2x = 10 |: (- 2)
 −2x −2
=
 10 −2

x = −5