Linear Equations with Different Kinds of Solutions

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Presentation transcript:

Linear Equations with Different Kinds of Solutions

Focus 9 - Learning Goal: The students will be able to set up and solve linear equations in one variable. 4 3 2 1 In addition to 3, student will be able to go above and beyond by applying what they know about solving linear equations in one variable. The student understands how to solve linear equations in one variable including equations whose solutions require expanding expressions using the distributive property and collecting like-terms. With no help the student has a partial understanding of how to solve linear equations in one variable. With help, the student may have a partial understanding of how to solve linear equations in one variable. Even with help, the student is unable to solve linear equations in one variable.

Solve: 11 + 3x – 7 = 6x + 5 – 3x First, combine like terms. 4 = 5 Is this true? Does 4 really equal 5? Then the “solution” is “no solution.” Next, subtract 3x from both sides.

Solve: 6x + 5 – 2x = 4 + 4x + 1 First combine like terms: 5 = 5 Is this true? Does 5 really equal 5? This means that you could put ANY number in for x and it would work. It has MANY solutions. Next, subtract 4x from both sides.

Try it: Substitute in 1, then 6, then 10 for “x” and see if it works. 4x + 5 = 5 + 4x 4(1) + 5 = 5 + 4(1) 4 + 5 = 5 + 4 9 = 9 True x = 6 4x + 5 = 5 + 4x 4(6) + 5 = 5 + 4(6) 24 + 5 = 5 + 24 29 = 29 True x = 10 4x + 5 = 5 + 4x 4(10) + 5 = 5 + 4(10) 40 + 5 = 5 + 40 45 = 45 True There are Infinitely Many Solutions to this equation.

Solve: 2(3 – 6m) = -30 First, use the distributive property. -6 -6 -12m = -36 -12 -12 m = 3 There is only ONE solution to this equation. Multiply 3 and (-6m) by 2. Subtract 6 from both sides. Divide by -12.

How many solutions can an equation have? x + 4 = 7 -4 -4 x = 3 One Solution x + 1 = x -x -x 1 = 0 No Solution x + 2 = 2 + x -x -x 2 = 2 Many Solutions