☰ Contents · Algebra

Solving first-degree equations in one unknown

Lessons 9 · 1 lessons · Sh.A. Alimov, O.R. Xolmuhamedov, M.A. Mirzaahmedov. Algebra, Grade 7, revised and expanded 5th edition. O‘qituvchi NMIU, Tashkent, 2017
9

Solving first-degree equations in one unknown

Textbook: pp. 38–43
GoalSolve first-degree equations in one unknown using the properties of equations.
New words
moving a term · hadni o‘tkazishcommon denominator · umumiy maxrajcoefficient · koeffitsiyentproperty of an equation · tenglama xossasi
Explanation

Two properties are basic when solving an equation: 1) any term can be moved from one side to the other if its sign is reversed; 2) both sides can be multiplied or divided by the same non-zero number. The procedure is: move the terms containing the unknown to the left and the rest to the right; combine like terms; divide both sides by the coefficient of the unknown. If there are brackets, open them first; if there are fractions, multiply both sides by the common denominator. If after combining we get 0 · x = b with b ≠ 0, there is no root; if we get 0 · x = 0, every value of x is a root. It is useful to substitute the result into the original equation.

Worked examples
7x - 15 = 3x + 9 → 7x - 3x = 9 + 15 → 4x = 24 → x = 6.
(x + 3) : 2 - (2x - 1) : 5 = 3. Multiply both sides by 10: 5(x + 3) - 2(2x - 1) = 30; 5x + 15 - 4x + 2 = 30; x + 17 = 30; x = 13.
3(x + 1) = 3x + 7: 3x + 3 = 3x + 7, 0 · x = 4. This equation has no root.
Class activity

“The balance”: an equation is explained as a balance scale. Expressions are written on its two pans; pupils apply the same operation to both pans (subtract, divide) to keep the balance and set x “free”.

Practice
1
Solve 6x - 9 = x + 16.
2
Solve 3(x - 2) - 2(x + 1) = 7.
3
Solve (2x - 1) : 3 = (x + 4) : 2.
4
Why is every value of x a root of 2(x + 1) = 2x + 2?