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Task 2.6 Solving Systems of Equations
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Solving Systems using Substitution Solve using Substitution if one variable is isolated!!! Substitute the isolated variable’s value into the 2 nd equation. Should have only one variable in the problem now Solve for the remaining variable Plug the value of the solved for variable back into one of the original equations to solve for remaining variable. Example: y= 3x + 2 2x + 2y = 20 2x + 2(3x + 2) = 20 2x + 6x + 4 = 20 8x = 16 X = 2 Then plug back in Y = 3(2)+ 2 Y = 6 + 2 Y = 8 Solution (2,8)
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Elimination Method With Same Coefficients Use elimination if both equations are in standard form Find a term from each equation with the same variable with the same coefficient. If the terms have the same sign we subtract one equation from the other. If the terms have different signs we add one equation to another. Once we have added or subtracted, we now have one variable that we can solve for. After solving for one variable, plug it into the original equation to solve for the other. Example: 4x + 3y = 11 2x + 3y = 13 2x = -2 X = -1 4(-1) + 3y = 11 -4 + 3y = 11 3y = 15 Y = 5 Solution (-1, 5)
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Elimination Method Without Same Coefficients Pick a variable to eliminate. Multiply the entire equation(s) by a common multiple of the chosen variable(similar to finding common denominators with adding and subtracting fractions) that we can eliminate. Find a term from each equation with the same variable with the same coefficient. If the terms have the same sign we subtract one equation from the other. If the terms have different signs we add one equation to another. Once we have added or subtracted, we now have one variable that we can solve for. After solving for one variable, plug it into the original equation to solve for the other. Example: 4x + 3y = 8 3x – 5y = -23 3(4x + 3y = 8) 4(3x – 5y = -23) 12x + 9y = 24 12x -20y = -92 29y = 116 Y = 4 4x + 3(4) = 8 4x + 12 = 8 4x = -4 X = -1 Solution (-1, 4)
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