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Algebra 1 Section 7.2.

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Presentation on theme: "Algebra 1 Section 7.2."— Presentation transcript:

1 Algebra 1 Section 7.2

2 Solving a System by Graphing
This works well only if the coordinates of the solution are small integers. Another method should be used if the solutions contain large integers or fractions.

3 Substitution Method This is used to find a single variable equation by replacing one of the variables with an equivalent expression.

4 Substitution Method Once this equation is solved, the result is substituted into one of the equations to find the value of the other variable.

5 Example 1 x + y = 5 y = -3x + 7 x + (-3x + 7) = 5 -2x + 7 = 5 -2x = -2

6 Example 1 Since we know that x = 1... y = -3x + 7 y = -3(1) + 7 y = 4
Solution: (1, 4)

7 Example 2 x + y = -4 x – y = 10 x = -y – 4 (-y – 4) – y = 10

8 Example 2 Since we know that y = -7... x = -y – 4 x = -(-7) – 4 x = 3
Solution: (3, -7)

9 Solving a System by Substitution
Solve either equation for either variable. Choose the easiest variable to solve for. Substitute the solution into the other equation to obtain an equation in one variable; then solve.

10 Solving a System by Substitution
Use the value of the known variable to solve for the other variable. Write the solution as an ordered pair.

11 Solving a System by Substitution
Check the solution by substituting it into each equation.

12 Example 3 Note here that the “easiest” variable to solve for is x in the first equation: x = 4y. 3(4y) + 2y = 8

13 Example 4 You need to write two equations involving l, the length, and w, the width. We are told the length is 6 ft longer than the width. l = w + 6

14 Example 4 The perimeter formula helps us to get the other equation:
2l + 2w = 84 Since we know that l = w 2(w + 6) + 2w = 84

15 Example 4 2(w + 6) + 2w = 84 2w + 12 + 2w = 84 4w + 12 = 84 4w = 72
l = w + 6 l = l = 24 width = 18 ft length = 24 ft

16 Homework: pp


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