## Simultaneous Equations

Aim: By the end of this chapter, you will be able to solve simultaneous equations with elimination and substitution methods.

1. The values of x and y that simultaneously satisfy the two linear equations, such as

y - 2x = 2
x + y = 3

can be found by employing either the elimination or the substitution methods.

 Elimination Method The method involves eliminating one of the variable from the two equations and solve for the other. You could choose either x or y to be eliminated. The choice however is often didacted by the ease of calculation. Let us work through this method using the two linear equations above. We will name the first equation (1) and the other (2). Example 1: y - 2x = 2 ; (1) y + x = 3 ; (2) Both equations have ONLY ONE y. It is thus obvious that y should be eliminated first. We could eliminate y by simplying letting (1) - (2):   [ y - 2x = 2 ] - [ y + x = 3 ]       0 - 3x = -1           x = 1/3 ; (3) Now we can substitute (3) into any of the original equations. We could choose (2) for its simplicity: y + (1/3) = 3 y = 8/3 Solution: x = 1/3 and y = 8/3. Example 2: 5x + 2y = 10 ; (1) 8x + 2y = 16 ; (2) (1) - (2) to eliminate 2y.   [ 5x + 2y = 10 ]; (1) - [ 8x + 2y = 16 ] ; (2)       -3x + 0 = -6           x = -6/-3           x = 2 ; (3) Substitute (3) into (1). 5(2) + 2y = 10 2y = 10 - 10 2y = 0 y = 0. Solution: x = 2 and y =0. Example 3: - 5x + 2y = 10 ; (1) 4x - 3y = 12 ; (2) Here, we can either eliminate x or y. Let us eliminate y. We need to multiply equation (1) by -3 (since the y in (2) has coefficient -3) and equation (2) by 2 to eliminate y. 15x - 6y = -30 ; (3) 8x - 6y = 24 ; (4) (3) - (4) to eliminate y.   [ 15x - 6y = -30 ]; (1) - [ 8x - 6y = 24 ] ; (2) 7x + 0 = -54             x = -54/7; (3) Substitute (3) into (2). We could also use any of the equations above and not necessary (2). 4(-54/7) - 3y = 12 - 3y = 12 + 216/7 -3y = 300/7 y = -100/7. Solution: x = -54/7 and y = -100/7. Substitution Method. The method involves making one of the variable the subject of the other variable. We could either choose either x or y from any one equation to be our subject. This new expression that we form out of, say, equation (2) is then substituted into equation (1). We will solve equation (1) for the remaining variable. Let us study this method using an example. Example 4: Le us use the same Example 1 in the Elimination Method. y - 2x = 2 ; (1) y + x = 3 ; (2) Equation (2) can be rewritten as y = 3 - x ; (3). Substitute (3) into (1): ( 3- x) - 2x = 2 3 - 3x = 2 1 = 3x x = 1/3 ; (4). Substitute (4) into (3): y = 3 - 1/3 y = 8/3. Solution: x = 1/3 and y = 8/3. Example 5: 6x + 4y = 24 ; (1) 5x - y = -5 ; (2) Equation (2) can be rewritten as y = 5x + 5 ; (3). Substitute (3) into (1): 6x + 4(5x + 5) = 24 6x + 20x + 20 = 24 26x = 4 x = 2/23 ; (4). Substitute (4) into (3): y = 5(2/23) + 5 y = 10/23 + 115/23. y = 125/23 Solution: x = 2/23 and y = 125/23. Example 6: 3y + 2x= 6 ; (1) -2y + 3x = 18 ; (2) For simplicity sake, we will choose (2) and make x the subject; x = (18+2y)/3 ; (3). Substitute (3) into (1): 3y - 2( 18 + 2y) /3 = 6 3y - 2(18)/3 - 2(2y)/3 = 6 3y - 12 - 4y/3 = 6 5y/3 = 18 y = 18(3)/5 y = 54/5 ; (4) Substitute (4) into (3): x = ( 18 + 2(54/5) )/3 x = 198/5 Solution: x = 198/5 and y = 54/5.
2. Personally I like the elimination method but you are free to form your own opinion. Both methods will yield identical solution.
3. Given two linear equations we could also plot these graphs and identify the intersection graphically. The point of intersection gives us the solution (x,y).
4. Examples:
Note: The intersection is obvious in example 2. There is no need to calculate anything if we have started our solution by plotting a graph.