Algebra 2 pt 17 chapter 1

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Substitution Method is often easier if one of the equations already has one variable expressed in terms of the other. That first variable can then be eliminated from the other equation by substitution. Example 1.28 says, "Solve the system{ 5x - 2y = 16 and y = x - 5" Solution The second equation gives y in terms of x.. In words, the second equation says "y is the same as x - 5." The expression (x - 5) can be substituted in the first equation, leaving an equation with only one variable. 5x -2y = 16 to 5x -2(x - 5) = 16 note the parentheses around x - 5. ^ y = x - 5 5x - 2x + 10 = 16 3x = 6 x = 2 to y = 2 - 5 = - 3 The solution is (2,_ 3) Any system of two linear equations may be solved by either method. In Example 1.29 Solve the system {2x = 3y - 14 and x + 3y = 2 solution 1 Use elimination. Rearrange the first equation into general form. Then eliminate y by addition.. 2x - 3y = - 14 x + 3y = 2 or - 4 + 3y = 2 ------------------- 3x = - 12 3y = 6 x = - 4 y = 2 solution 2 Use substitution. 2x =3y - 14 to 2x = 3y - 14 to 2(-3y + 2) = 3y - 14 x + 3y = 2 to x = - 3y +2 -6y +4 = 3y- 14 - 9y = - 18 y =2 to x = - 3(2) + 2 = - 4 Solution for both calculations were: (-4,2) The book goes into graphs from here. The 'Graphical Solutions will be presented on next Wednesday.