## 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.
 A TI-83 will be useful here. Method 1: Graphing We start by presing [ Y = ] (top right). This will bring up a screen in which we will enter equation (1) into Y1 and equation (2) into Y2. Make sure you set the [ WINDOW ] (top second right) in a way that you will be able to see the intersection, if there is one. Press [ GRAPH ] (top left). This will plot your graphs. Press [ 2nd ] [ TRACE ] to obtain the CALC function that brings us to a page called "CALCULATE". Scroll down to " 5:intersect " and press [ ENTER ] . The first prompt is "First curve?" and there will be a blinker on the first curve Y1. The equation Y1 is displayed on the upper left. Make sure that this is correct. If it is correct then press [ ENTER ] . It will then ask for the "Second curve?" and again make sure that it is Y2. If it is correct then press [ ENTER ] . The last prompt is "Guess?". Moves your blinker to the intersection and press [ ENTER ]. The intersection point is reported at the bottom of the screen. Method 2: Matrix Method For illustration, we going to use a previous example of - 5x + 2y = 10 ; (1) 4x - 3y = 12 ; (2) For this technique to work, always arrange your equations with x first and then y as we had done here. [You could have y first and then x but not one equation with x first and the second equation with y first.] Press [2nd][x-1] for Matrix and move to EDIT as in Screen 1. The dimension of the matrix is 2x3 ("2 by 3") where 2 stands for 2 equations and 3 terms in our case. The number 10 in equation 1 is taken as a term. Enter the information as in Screen 2. Press[2nd][MODE] to quit from the matrix entry. Press [2nd][x-1] for Matrix again and move to Math and B:rref as in Screen 3. Press [ENTER] [2nd][x-1] and at Names select 1:[A] and [ENTER]. We will now obtain the top three lines in Screen 4. It is a good practice to change the answers back to fraction. Thus, press [MATH] , select 1:>Frac [ENTER] to obtain the results in Screen 4. We will report the answer as x = -54/7 and y = -100/7. The rref stands for reduced row-echelon form but for our purpose this is a symbol that helps us to solve simultaneous equations. This method can also be used for 3 equations with 3 variables. In general, this method works for n equations with n variables. Method 3: Use the Application Polysmlt from Ti. This application is acceptable for IB examination and can be downloaded from here.

Special Cases

1. Both the elimination and substitution methods are useful when the given two lines actually intersect.
2. Consider x + y = 5 and x + y = 10 .
If you plot these lines then you will observe that one is parallel to the other. They do not intersect each other. The is NO Real Number Solution to this system. Given such a system, we should not waste our time employing either the elimination or the substitution method.
3. Consider x + y = 5 and 3x + 3y = 15 .
Observe that the second equation (2) is a multiple of the first(1), namely three. If you actually plot these two lines, you will find (2) is perfectly on top of (1). You will effectively see only one line. Thus, every point on (2) intersects with (1). We have infinitely many solutions . Any x and its associated y in the domain is a solution. Both elimination and substitution methods are not useful here.
4. Plot graphs for the two systems above to convince yourself that the former are parallel lines and the latter is essentially one line.