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Solving Mixed Equations. Grade Seven & Eight Mathematics M. M. Couturier. Solving Equations. In this class, we will combine the skills and techniques that we learned in the previous classes; how to solve for examples like: x + 6 = 10 and 3x = 15.
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Solving Mixed Equations Grade Seven & Eight Mathematics M. M. Couturier
Solving Equations • In this class, we will combine the skills and techniques that we learned in the previous classes; how to solve for examples like: • x + 6 = 10 and • 3x = 15. • However, this time we will combine addition/subtraction and multiplication/division situations.
Solving Equations • Let’s look at a few examples; solve for x: • 2x + 5 = 25 • See if you can solve for this!
Solving Equations • Let’s look at a few examples; solve for x: • 2x + 5 = 25 • Let’s isolate the variable by using reverse BEDMAS or SAMDEB; first we get rid of the +5. • 2x + 5 - 5 = 25 - 5 • 2x = 20 • Now we get rid of the multiple of 2. • 2x = 20 • 2 2 • x = 10; Therefore; x = 10.
Solving Equations • Example # 2; solve for x in • 6x - 14 = 46 • Try this on your own for now!
Solving Equations • Example # 2; solve for x in • 6x - 14 = 46 • Isolate the variable first getting rid of the -14. • 6x - 14 + 14 = 46 + 14 • 6x = 60 • Now get rid of the multiple of 6. • 6x = 60 • 6 6 • x = 10; Therefore; x = 10.
Solving Equations • Example # 3; solve for x in; • x - 8 = 10 • 2 • Try this on your own!!!
Solving Equations • Example # 3; solve for x in; • x - 8 = 10 • 2 • Isolate the variable by getting rid of the -8. • x - 8 + 8 = 10 + 8 • 2 • x = 18 • 2
Solving Equations • Now we get rid of ½ by multiplying by 2. • 2x = (2)18 • 2 • x = 36 • Therefore, x = 36.
Solving Equations • Example # 4; solve for x in; • 5x - 5 = 5 • 4 4 • Try this one on your own!!!
Solving Equations • Example # 4; solve for x in; • 5x - 5 = 5 • 4 4 • Now there is a secret that we can use. If you don’t like dealing with fractions, then multiply EVERYTHING by the largest denominator, which in this case is 4. • (4)5x - (4)5 = (4)5 • 4 4
Solving Equations • 5x - 5 = 20 • Now we have a fractionless equation: • 5x - 5 + 5 = 20 + 5 • 5x = 25 • 5x = 25 • 5 5 • x = 5; Therefore, x = 5.
Solving Equations • Try a few of these: • A) 4x + 15 = 35 • B) -2x – 6 = -14 • C) 7x –1 = 20 • D) 5x – 50 = 20
Solving Equations • Solution A) • 4x + 15 = 35 • 4x + 15 – 15 = 35 – 15 • 4x = 20 • 4x = 20 • 4 4 • x = 5; Therefore, x = 5.
Solving Equations • Solution for B) • -2x – 6 = -14 • -2x – 6 + 6 = -14 + 6 • -2x = -8 • -2x = -8 • -2 -2 • x = 4; Therefore, x = 4.
Solving Equations • Solution for C) • 7x –1 = 20 • 7x - 1+ 1 = 20 + 1 • 7x = 21 • 7x = 21 • 7 7 • x = 3; Therefore, x = 3.
Solving Equations • Solution for D) • 5x – 50 = 20 • 5x – 50 + 50 = 20 + 50 • 5x = 70 • 5x = 70 • 5 5 • x = 14; Therefore, x = 14.