* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Solve Systems of Equations by the Substitution Method
Two-body Dirac equations wikipedia , lookup
Maxwell's equations wikipedia , lookup
Two-body problem in general relativity wikipedia , lookup
Debye–Hückel equation wikipedia , lookup
Schrödinger equation wikipedia , lookup
Unification (computer science) wikipedia , lookup
BKL singularity wikipedia , lookup
Perturbation theory wikipedia , lookup
Dirac equation wikipedia , lookup
Van der Waals equation wikipedia , lookup
Euler equations (fluid dynamics) wikipedia , lookup
Navier–Stokes equations wikipedia , lookup
Equations of motion wikipedia , lookup
Calculus of variations wikipedia , lookup
Equation of state wikipedia , lookup
Itô diffusion wikipedia , lookup
Derivation of the Navier–Stokes equations wikipedia , lookup
Computational electromagnetics wikipedia , lookup
Differential equation wikipedia , lookup
Schwarzschild geodesics wikipedia , lookup
Solve Systems of Equations by the Substitution Method Solving a system of equations by graphing has several limitations. First, it requires the graph to be perfectly drawn – if the lines are not straight, we may arrive at the wrong answer. Second, graphing is not a great method to use if the answer is really large or if the answer is a decimal or fraction. For these reasons, graphing is rarely used to solve systems. The first algebraic approach, discussed in this section, is called the substitution method. When the value of one variable is known, it can be substituted into the other equation. Example 1: Solve the system: x = 5 y = 2x – 3 y = 2(5) – 3 y = 10 – 3 y=7 Substitute 5 for x into the second equation The solution to this system is (5, 7) When one variable equals an expression, that expression can also be substituted into the other equation. It is important to use parenthesis around the substituted expression to lessen the chance of making a mistake when distributing. Once that equation is solved and a value is known for one of the variable, we proceed as in Example 1. Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) Example 2: Solve the system: 2x – 3y = 7 y = 3x – 7 2x – 3(3x – 7) = 7 2x – 9x + 21 = 7 - 7x + 21 = 7 - 21 - 21 - 7x = -14 -7 -7 x=2 y = 3(2) – 7 y=6–7 y = -1 Substitute (3x – 7) into the first equation Distribute -3 Combine like terms Subtract 21 from both sides Divide both sides by -7 Substitute 2 into the second equation The solution to this system is (2, -1) To check the answer: 2(2) – 3(-1) = 7 4+3=7 7=7 True -1 = 3(2) – 7 -1 = 6 – 7 -1 = -1 True Since (2, -1) satisfies both equations, it is the solution to the system Sometimes the lone variable (variable without a coefficient) is not alone on one side of the equation. In this case, we isolate it by solving for the lone variable. Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) Example 3: Solve the system: 3x + 2y = 1 x – 5y = 6 x – 5y = 6 + 5y + 5y x = 6 + 5y 3(6 + 5y) + 2y = 1 18 + 15y + 2y = 1 18 + 17y = 1 - 18 - 18 17y = -17 17 17 y = -1 Add 5y to both sides to isolate x Substitute (6 + 5y) into the first equation Distribute 3 Combine like terms Subtract 18 from both sides x = 6 + 5(-1) Substitute -1 into the second equation x=6–5 x=1 The solution to this system is (1, -1) To check the answer: 3(1) + 2(-1) = 1 3–2=1 1=1 True 1 – 5(-1) = 6 1+5=6 6=6 True Since (1, -1) satisfies both equations, it is the solution to the system Just as with graphing, it is possible to have no solution (parallel lines) or infinite solutions (same line) when solving algebraically. Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) Example 4: Solve the system: y = 3x – 4 2y – 6x = -8 2(3x – 4) – 6x = -8 Substitute (3x + 4) into the second equation 6x – 8 – 6x = -8 -8 = -8 Combine like term & the variables are gone This is a true statement, so there are infinite solutions, meaning the two equations are equivalent and represent the same line. Example 5: Solve the system: 6x – 3y = -9 y = 2x + 5 6x – 3(2x + 5) = -9 Substitute (2x + 5) into the first equation 6x – 6x – 15 = -9 -15 = -9 Combine like terms & the variables are gone This is a false statement, so there is no solution, meaning the two equation represent two different parallel lines. If there is no lone variable, we select one variable in one of the equations to isolate. In this situation, it is a good possibility that the resulting equation will contain fractions. Example 6: Solve the system: 5x – 6y = -14 -3x + 4y = 12 -3x + 4y = 12 +3x +3x 4y = 12 + 3x 4 4 4 Add 3x to both sides Divide each term by 4 3 y=3+ x 4 Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) 3 3 5x – 6(3 + x) = -14 Substitute (3 + x) into the first equation 5x – 18 – x = -14 Distribute 6 x = -14) Clear the fraction by multiplying by LCD 2(5x – 18 – 4 9 2 9 2 10x – 36 – 9x = -28 x – 36 = -28 +36 +36 x =8 -3(8) + 4y = 12 -24 + 4y = 12 +24 +24 4y = 36 4 4 y=9 4 Add 36 to both sides Substitute 8 into the second equation Add 24 to both sides Divide both sides by 4 The solution to this system is (8, 9) To check the answer: 5(8) – 6(9) = -14 40 – 54 = -14 -14 = -14 True -3(8) + 4(9) = 12 -24 + 36 = 12 12 = 12 True Since (8, 9) satisfies both equations, it is the solution to the system When two equations equal the same value, they must be equivalent to each other. Basically we set the equations equal to each other and solve for the remaining variable. Then we can substitute the resulting into either equation. Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0) 3 Example 7: Solve the system: y = x + 2 4 2 y= x+1 3 3 2 4 3 3 2 4 3 x+2= x+1 Both equations equal y, so they equal each other 12( x + 2 = x + 1) Multiply by the LCD to clear the fractions 9x + 24 = 8x + 12 - 8x - 8x x + 24 = 12 - 24 - 24 x = -12 3 Subtract 8x from both sides Subtract 24 from both sides y = (-12) + 2 Substitute -12 into the first equation y = -9 + 2 y = -7 The solution to this system is (-12, -7) 4 3 To check the answer: -7 = (-12) + 2 4 -7 = -9 +2 -7 = -7 True 2 -7 = (-12) + 1 3 -7 = -8 + 1 -7 = -7 True Since (-12, -7) satisfies both equations, it is the solution to the system Modified from Beginning and Intermediate Algebra, by Tyler Wallace, CC-BY 2010. Licensed under a Creative Commons Attribution 3.0 Unported License (http://creativecommons.org/licenses/by/3.0)