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TYPES OF SOLUTIONS OF LINEAR EQUATIONS

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Presentation on theme: "TYPES OF SOLUTIONS OF LINEAR EQUATIONS"— Presentation transcript:

1 TYPES OF SOLUTIONS OF LINEAR EQUATIONS
Algebraic Form Number of Solutions Description a = b None There are no values of the variable for which the equation is true. x = a One The equation is true for exactly one value of the variable. a = a Infinitely many The equation is true for all values of the variable.

2 1. 2x – 4 = -x - 1 + x + x 3x – 4 = - 1 + 4 + 4 3x = 3 3 = 3 x = 1
Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. x – 4 = -x - 1 + x x Add 1x to each side Simplify Add 4 to each side Divide both sides by 3 3x – 4 = - 1 3x = 3 = 3 x = 1 The result is an equation of the form x = a. This equation is true for exactly one value. So, the equation has one solution.

3 Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. x – 4 = 2(x – 2) Distributive Property Subtract 2x from both sides Simplify 2x – 4 = 2x – 4 - 2x x – 4 = - 4 The result is an equation of the form a = a. This equation is true for all values of x. So, the equation has infinitely many solutions.

4 Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. x – 4 = 2(x + 1) Distributive Property Subtract 2x from both sides Simplify 2x – 4 = 2x + 2 - 2x x – 4 = 2 The result is an equation of the form a = b. There are no values of x for which the equation is true. So, the equation no solution.

5 a. 5x + 8 = 5(x + 3) b. 9x = 8 + 5x - 5x - 5x 5x + 8 = 5x + 15
Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. a x + 8 = 5(x + 3) b x = 8 + 5x - 5x x 5x + 8 = 5x + 15 - 5x x 4x = 8 8 = 15 x = 2 no solution form a = b one solution form x = a

6 c. 6x + 12 = 6(x + 2) d. 7x - 11 = 11 - 7x + 7x + 7x 6x + 12 = 6x + 12
Tell whether each equation has one solution, infinitely many solutions, or no solutions. Justify your answer. c x + 12 = 6(x + 2) d x - 11 = x + 7x x 6x + 12 = 6x + 12 - 6x x 14x - 11 = 11 12 = 12 14x = 22 x = = 1 4 7 infinitely many solutions form a = a one solution form x = a


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