Substitution and Elimination The Method of Substitution The

Substitution and Elimination

The Method of Substitution

The Method of Substitution To solve a system of two equations in two variables by substitution, follow these steps: ü Solve one of the equations for one variable in terms of the other variable. ü Substitute the expression found in step 1 to the other equation to obtain an equation of one variable. ü Solve the equation obtained in step 2.

The Method of Substitution To solve a system of two equations in two variables by substitution, follow these steps: ü Substitute the solution from step 3 to the expression obtained in step 1 to find the value of the other variable. ü Check the solution to ensure that it satisfies both of the original equations.

The Method of Substitution Example 1: Solve the following systems of linear equations using substitution method. -2 x + y = 1 x – 3 y = 2 equation 1 equation 2

The Method of Substitution Solution: Begin by solving for x in the second equation. x = 3 y + 2 (solve for x in equation 2) Next, substitute the solve expression for x in equation 1. -2 (3 y + 2) + y = 1 -6 y – 4 + y = 1 -5 y = 5 y = -1

The Method of Substitution Solution: To find the value of x, substitute the value of y to equation 2. x – 3 y = 2 x – 3(-1) = 2 x = -3 + 2 x = -1 • Thus, the solution is (-1, -1). Check to see that it satisfies both original equations

The Method of Substitution Example 2: Solve the system : 3 x + 2 y = 11 y = 4 x (1) (2)

The Method of Substitution Solution: Substitute 4 x for y in equation (1). 3 x + 2(4 x) = 11 3 x + 8 x = 11 11 x = 11 x = 1 Substitute 1 for 4 x in equation (2) y = 4(1) y = 4 • Thus, the solution is (1, 4). Check to see that it satisfies both original equations

The Method of Substitution Example 3: Solve the system : x + 2 y + 3 z = 9 2 x – y + z = 8 3 x -z = 3 (eq. 1) (eq. 2) (eq. 3)

The Method of Substitution Solution: Begin by solving for z in the third equation. Z = 3 x - 3 (solve for z in equation 3) (eq. 4) Substitute the solve expression for z in equation 1. x + 2 y + 3 z = 9 x + 2 y + 3(3 x – 3) = 9 x + 2 y + 9 x – 9 = 9 2 y + 10 x = 18 2 y = -10 x + 18 y = -5 x + 9 (eq. 5)

The Method of Substitution Solution: Substitute equation 4 and 5 in equation 2. 2 x – y + z = 8 2 x – (-5 x + 9) + (3 x – 3) = 8 2 x + 5 x -9 + 3 x – 3 x = 8 10 x – 12 = 8 10 x = 8 + 12 10 x = 20 x= 2

The Method of Substitution Solution: Substitute the value of x to equation 3. 3 x – z = 3 3 (2) – z =3 6 – z =3 -z = 3 – 6 -z = - 3 Z = 3

The Method of Substitution Solution: Substitute the value of x and z to equation 1. x = 2 y = 3 z = 9 (2) + 2 y + 3 (3) = 9 2 + 2 y + 9 = 9 2 y + 11 =9 2 y = 9 – 11 2 y = -2 Y = -1 • The value of x=2; y= -1; and z=3

The Method of Substitution Example 4: Solve the system : x + 2 y – z = 3 2 x + y – 3 z = 1 x + 8 y – 3 z = -7 (eq. 1) (eq. 2) (eq. 3)

The Method of Substitution Solution: Begin by solving for y in the 2 nd equation. Y = -2 x + 3 z + 1 (eq. 4) Substitute (eq. 4) in equation 1. x + 2 y – z = 3 x + 2(-2 x + 3 z + 1) – z = 3 x -4 x + 6 z + 2 – z = 3 -3 x + 5 z = 1 (eq. 5)

The Method of Substitution Solution: Substitute (eq. 4) in equation 3. x + 8 y – 3 z = -7 x + 8(-2 x + 3 z + 1) – 3 z = -7 x -16 x + 24 z + 8 – 3 z = -7 -15 x + 21 z + 8 = -7 -15 x + 21 z = -7 – 8 -15 x + 21 z = -15 (eq. 6)

The Method of Substitution Solution:

The Method of Substitution Solution: Substitute the value of z in equation 5. -3 x + 5 z = 1 -3 x + 5(5) = 1 -3 x + 25 = 1 -3 x = -24 x = 8 Substitute the value of z and x in equation 1. x + 2 y – z = 3 8 + 2 y – 5 = 3 2 y = 0 • The value of x=8; y= 0; and z=5

The Method of Elimination

The Method of Elimination The Method Of Elimination A third and often the easiest method of solving a system of equations is the elimination method. The objective of this process is to obtain an equation containing only one variable.

The Method of Elimination Example 1: Solve the following system of linear equations using elimination method. x+y=8 x–y=4 equation 1 equation 2

The Method of Elimination Solution: Note that one equation contains +y and the other contains –y. Adding the equations will eliminate the variable y. thus, obtaining one equation containing only one unknown, x. x+y=8 x–y=4 2 x = 12 Solve the remaining variable, x. 2 x = 12 x = 6

The Method of Elimination Solution: Solve for y by substituting 6 for x in either of the original equations. Using eq. 1. x+y=8 6+y=8 y = 2 • The solution is (6, 2).

The Method of Elimination To use the method of elimination to solve a system of two linear equations in x and y, follow these steps: ü If necessary, rewrite each equation in standard form, that is, the terms containing variables are on the left side of the equal sign and the constant on the right side. ü Obtain coefficients for x (or y) that differ only in sign by multiplying all terms of one or both equations by suitable constants.

The Method of Elimination To use the method of elimination to solve a system of two linear equations in x and y, follow these steps: ü Add the equations to eliminate one variable and solve for the resulting equation. ü Substitute the value obtained in step 3 to either of the original equations and solve for the other variable. ü Check your solutions in both of the original equations.

The Method of Elimination Example 2: Solve the following system of linear equations. 2 x + 5 y = 3 equation 1 5 x – 3 y = 23 equation 2

The Method of Elimination Solution: To obtain coefficients of x that differ only in sign, multiply equation 1 by 5 and equation 2 by -2. 2 x + 5 y = 3 5 x – 3 y = 23 10 x + 25 y = 15 -10 x + 6 y = -46 31 y = -1 (multiply eq. 1 by 5) (multiply eq. 2 by -2) (add equations) (divide both sides by 31)

The Method of Elimination Solution: To solve for the variable x, substitute y = -1 to either of the original equations. 2 x + 5(-1) = 3 2 x – 5 = 3 2 x = 8 x = 4 • Therefore, the solution is (4, -1)

The Method of Elimination Example 3: Solve the system : x+y=8 x–y=2 (1) (2)

The Method of Elimination Solution: (x + y) + (x – y) = 8 + 2 2 x = 10 x= 5 (addition property of equality) (combine like terms) (simplify) Take equation (1), x+ y = 8 5+y=8 y = 3 (substitute) (simplify ) • The solution is ordered pair (5, 3)

The Method of Elimination Example 4: Solve the system : x + 2 y – z = 3 2 x + y – 3 z = 1 x + 8 y – 3 z = -7 (eq. 1) (eq. 2) (eq. 3)

The Method of Elimination Solution: Eliminate x. Add eq. (1) and eq. (3). Multiply (eq. 1) by -1. x + 2 y – z = 3 -x – 2 y + z = -3 x + 8 y – 3 z = -7 6 y – 2 z = -10 (eq. 4) Eliminate x. Add eq. (1) and eq. (2). Multiply (eq. 1) by -2. -2 ( x + 2 y – z = 3 ) -2 x – 4 y + 2 z = -6 2 x + y – 3 z = 1 -3 y – z = -5 (eq. 5)

The Method of Elimination Solution: Solve for z using eq. (4) and eq. (5). Multiply eq. (5) by 2. 2(-3 y – z = -5) 6 y – 2 z = -10 -4 z = -20 z = 5 Solve for y by substituting the value of z to eq. (4) 6 y – 2 z = -10 6 y – 2(5) = -10 6 y – 10 = -10 6 y = 0

The Method of Elimination Solution: Solve for x by substituting the value of z and y to eq. (1) x + 2 y – z = 3 x + 2(0) – (5) = 3 x– 5=3 x=3+5 x = 8 • The solution are x=8, y=2, and z=5

The Method of Elimination Example 5: Solve the system : x + 2 y + 3 z = 9 2 x – y + z = 8 3 x -z = 3 (eq. 1) (eq. 2) (eq. 3)

The Method of Elimination Solution: Eliminate x. Add eq. (1) and eq. (2). Multiply (eq. 1) by -2. x + 2 y + 3 z = 9 -2 x – 4 y -6 z = -18 2 x – y + z = 8 -5 y – 5 z = -10 y + z = -2 (eq. 4) Eliminate x. Add eq. (1) and eq. (3). Multiply (eq. 1) by -3. x + 2 y + 3 z = 9 -3 x – 6 y - 9 z = -18 3 x -z = 3 -6 y -10 z = -24 3 y + 5 z = 12 (eq. 5)

The Method of Elimination Solution: Solve for z using eq. (4) and eq. (5). Multiply eq. (4) by -3. y + z = -2 -3 y – 3 z = -6 3 y + 5 z = 12 2 z = 6 z = 3 Solve for y by substituting the value of z to eq. (3) 3 x – z = 3 3 x – 3 = 3 3 x = 6 x = 2

The Method of Elimination Solution: Solve for x by substituting the value of z and x to eq. (1) x + 2 y + 3 z = 9 2 + 2 y – 3(3) = 9 2 y = 9 -9 -2 2 y = -2 y = -1 • The solution are x=2, y=1, and z=3

Applications

Applications APPLICATIONS When solving applied problem solving involving two variables, the initial work is always the same. We assign variables and write a system of equations. The only decision that remains concerns the choice of method for the system.

Applications Example 1: Glen Rose and James are planning to move and need to buy some cardboard boxes. At the bookstore, they can buy either 10 small boxes and 12 medium boxes for 29 php. Find the cost of each size box.

Applications Solution: Let: x = be the cost of the small box y = be the cost of the medium box 10 x + 12 y = 42 5 x + 10 y = 29 (equation 1) (equation 2) To solve this system of linear equations, you can use method of elimination. To obtain the coefficient of x that differ only in sign, multiply equation (2) by -2. 10 x + 12 y = 42 (5 x + 10 y = 29) (-2) 10 x + 12 y = 42 -10 x – 20 y = -58 -8 y = -16 y = 2

Applications Solution: Thus, the cost of each medium size box is 2 php. By substituting this value into equation 1, the cost of a small box can be solved. 10 x + 12(2) = 42 10 x + 24 = 42 10 x = 18 x = 1. 8 Thus, the cost of each small-size box is 1. 8 php and the cost of each medium-size box is 2 php.

Applications Example 2: Two cars starts out together and travel opposite directions. At the end of 3 hours, they are 345 kilometers apart. Find the speed of two cars, if one car travels 15 kph faster than other.

Applications Solution: Let: x = rate of first car y = rate of the second car Arrange the information in a chart. speed time distance First car x 3 3 x Second car Y 3 3 y

Applications Solution: The distance between the two cars is 345 km. 3 x + 3 y = 345 (eq. 1) The speed of the first car is 15 kph faster than the other. x = 15 + y (eq. 2) To solve this system of linear equation, substitution method can be used.

Applications Solution: Substitute equation 2 to equation 1. 3 x + 3 y = 345 3(15 + y) + 3 y = 345 45 + 6 y = 345 6 y = 300 y = 50 Thus, the rate of the second car is 50 kph. To solve for the variable x, back-substitute this value to the 2 nd equation. Therefore, the rate of the first car is X = 15+ y X = 15 + 50 X = 65 kph

Applications APPLICATIONS When solving a system of linear equations by either substitution or elimination, and arrived at an equation that is false (0 = constant), the system is inconsistent and has no solution. If the equation obtain is true (constant = constant), then the system is dependent and has infinitely many solutions.

Applications Example 3: Solve the following system of linear equations: 2 x + 3 y = 5 (equation 1) 4 x + 6 y = 9 (equation 2)

Applications Solution: To obtain coefficients of x that differ only in sign, multiply equation 1 by -2. 2 x + 3 y = 5 -4 x – 6 y = 10 (multiply eq. 1 by -2) 4 x + 6 y = 9 0 = -1 (false statement) Because there are no values of x and y for which 0 = -1, the system is inconsistent and has no solution.

Applications Example 4: Solve the following system of linear equations: 6 x – 8 y = 4 (equation 1) 3 x - 4 y = 2 (equation 2)

Applications Solution: To obtain coefficients of x that differ only in sign, multiply equation 2 by -2. 6 x – 8 y = 4 3 x - 4 y = 2 -6 x + 8 y = -4 (multiply eq. 2 by -2) 0 = 0 (add equations) Because the two equations turned out to be equivalent, the system has infinitely many solutions.

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