|
This Lesson (LINEAR EQUATIONS IN TWO DIMENSIONS) was created by by Theo(675)   About Theo:
This lesson covers an overview of LINEAR EQUATIONS
REFERENCES Slope of a Line (wtamu) Equations of Lines (wtamu) Graphing a Line (wtamu) Parallel and Perpendicular Lines (wtamu) Slopes and Equations of Lines (umsl) Slopes and Equations of Lines (regentsprep) Equations of Lines (cliffsnotes) Equation of a Line (analyzemath) Writing Equations of Lines (algebralab) Straight line Equations (purplemath) Parallel and perpendicular Lines(purplemath) Solving Simultaneous Linear Equations (the mathpage) Simultaneous Equations (thinkquest) Simultaneous Equations (mathsteacher) Solving Simultaneous Equations (gcsebitesize) Systems of Linear Equations (purplemath) DIMENSIONS A dimension is measurement of length in one direction. A one dimensional object could be represented by a line. It has length but no width (theoretically). A two dimensional object could be represented by a square It has length and width but no depth (theoretically). A three dimensional object could be represented by a cube. It has length and width and depth. Once you get past three dimensions, it becomes very difficult to visualize what the additional dimensions would look like. They are, however, still linear equations. Advanced techniques are used to solve them. These techniques could also be applied to two and three dimension linear equations but it is not necessary to do so. The techniques discussed here and in the lesson on linear equations in three dimensions are sufficient. This lesson concerns itself with two dimensions only. Any reference to dimensions other than two will be explicitly stated. POINTS ON A LINE Points on a line are in the form of (x,y) where x is the x coordinate of the point, and y is the y coordinate of the point. To find the point (x,y), imagine a vertical line from the x-coordinate on the x-axis, and a horizontal line from the y-coordinate on the y-axis. The intersection of these imaginary lines is the location of the point. If x is positive, the x-coordinate on the x-axis is x units to the right of the y-axis. If x is negative, the x-coordinate on the x-axis is x units to the left of the y-axis. If y is positive, the y-coordinate on the y-axis is y units above the x-axis. If y is negative, the y-coordinate on the y-axis is y units below the x-axis. A picture of the points (6,4), (0,0) and (-6,-4) with comments can be viewed by clicking on the following hyperlink. Points On a Line The vertical and horizontal line are shown as dashed lines to indicate they are invisible. They are only used as a guide to get you to the points that you want to locate. DEFINITION OF LINEAR EQUATION A linear equation is an equation of the first degree which means that the highest exponent in the equation is 1. The graph of a linear equation in two and three dimensions is a straight line. STANDARD FORM OF THE EQUATION FOR A STRAIGHT LINED The standard form of a linear equation is: ax + by = c where: a is the coefficient of the x term. b is the coefficient of the y term. c is a constant Example: 4x + 2y = 12 a = 4 b = 2 c = 12 A graph of this equation is shown below: SLOPE-INTERCEPT FORM OF THE EQUATION FOR A STRAIGHT LINE In order to graph this equation, we had to convert it to a form of y = f(x). Here's how we did that: Original equation is: 4x + 2y = 12 Subtract 4x from both sides of this equation to get: 2y = -4x + 12 Divide both sides of this equation by 2 to get: y = -2x + 6 The equation is now in a form that allows us to graph it. Please note that 2x and 2*x mean the same thing, which is 2 multiplied by x. Where it can be inferred easily, I leave the * symbol out. Where it is desired to make sure that multiplication is indicated, I insert it. The equation of y = -2x + 6 is the same as the equation y = -2*x + 6. y = -2x + 6 is also called the slope-intercept form of the equation. It is called that because this form of the equation tells us what the slope of the equation is, and what the y intercept of the equation is. The slope of this equation is equal to -2. The y intercept of this equation is equal to 6. The general form of the slope-intercept form of a linear equation is: y = mx + b where: m is the slope b is the y intercept The original equation in standard form of 4x = 2y = 12 was transformed to the slope-intercept form of y = -2x + 6 in order to graph it. An additional bonus is that we can find the slope and the y intercept directly from this form of the equation. DETERMINING THE SLOPE OF A LINE GIVEN 2 POINTS ON THE LINE If you are given 2 points on a line, then you can find the slope by using the following formula. Slope of a line = (y2-y1) / x2-x1) (x1,y1) is one point of the line. (x2,y2) is the other point of the line. Note that x1,y1 identifies a unique point separate from (x,y). Same with (x2,y2). Same with (x3,y3), etc.... They are numbered to show that they are unique points from each other, and from the general equation point of (x,y). The slope is the change in the value of y divided by the change in the value of x. A large change in the value of y and a small change in the value of x will provide a steep slope. A small change in the value of y and a large change in the value of x will provide a shallow slope. Example 1: Let the point (x1,y1) = (2,2) Let the point (x2,y2) = (3,6) Change in the value of y = (y2-y1) = 6-2 = 4 Change in the value of x = (x2-x1) = 3-2) = 1 Slope is 4/1 = 4 This would be a relatively steep slope. Example 2: Let the point (x3,y3) = (2,2) Let the point (x4,y4) = (10,3) Change in the value of y = (y4-y3) = 3-2 = 1 Change in the value of x = (x4-x3) = (10-2) = 8 Slope is 1/8 = .125 This would be a relatively shallow slope. A graph of these two lines is shown below. Note the y intercept of both lines. We will solve for the equation of these 2 lines and the intersection of these 2 lines right after the graph. DETERMINING THE Y-INTERCEPT OF A LINE AND THE X-INTERCEPT OF A LINE The first line has a relatively steep slope. It goes through the points (2,2) and (3,6) The slope-intercept form of a line is y = mx + b where m is the slope and b is the y-intercept. The y-intercept is the value of y when x = 0. The slope of the first line is 4 that we calculated just above. The slope-intercept form of the equation for this line becomes: y = 4x + b We need to solve for b. We do that by taking any point on the line and replacing x with the x-coordinate and y with the y-coordinate in the slope-intercept form of the equation and solving for b. We'll take the point (3,6). This was one of the points that we used when we solved for the slope of the line. Our equation of y = 4x + b becomes 6 = 4*3 + b becomes 6 = 12 + b after we replace y with 6 and x with 3. We subtract 12 from both sides of this equation to get: 6-12 = b which becomes: b = -6 The slope-intercept form of the equation for our first line is: y = 4x - 6 You can see on the graph of this equation that the line with the steep slope crosses the y-axis at the point (0,-6). We did the same with the second equation to get: y = (1/8)*x + (7/4) Our equation was y = (1/8)*x + b right after we solved for the slope. We then took the point (2,2) on that line and substituted for x and y in the equation to get: 2 = (1/8)*(2) + b We solved for b as follows: b = 2 - (1/8)*(2) which became: b = 2 - (2/8) which became: b = 1+(3/4) which became: b = (7/4) making our equation equal to y = (1/8)*x + (7/4). If you look at the equation, you will see that the second line with the shallow slope intercepts the y-axis at the point (0,(7/4)). POINT-SLOPE FORM OF THE EQUATION OF A STRAIGHT LINE There is another form of the equation of a line that is also used to find the y intercept. That form is called the point-slope form of the equation for a straight line and is shown as: (y-y1) = m*(x-x1) Note that the point is identified uniquely as (x1,y1) to distinguish it from the general point of (x,y) normally used in the equation. Once you know the slope, you then take any point on the line as before and replace (x1,y1) with it to get the point-slope form of the equation. Example: Given the point (3,6) and the slope of 4, you would plug these values into the point-slope form of the equation to get: y-6 = 4*(x-3) x1 was replaced with 3 and y1 was replaced with 6. You simplify this expression and solve for y as follows: y-6 = 4x - 12 Add 6 to both sides of this equation to get: y = 4x - 12 + 6 which becomes: y = 4x - 6 You have just derived the slope-intercept form of the equation for this line. Y-INTERCEPT OF THE EQUATION OF A LINE -6 is the y intercept because when you let x = 0, the equation becomes y = -6 The value of y when x = 0 is -6. The line formed by the equation of y = 4x-6 crosses the y-axis at y = -6. X-INTERCEPT OF THE EQUATION OF A LINE If you are asked to find when the line crosses the x-axis, you simply replace y with 0 and solve for x. Your equation of y = 4x - 6 becomes 0 = 4x - 6. Add 6 to both sides of this equation to get: 4x = 6 Divide both sides of this equation by 4 to get: x = 6/4 = 1.5 You have now found everything you need to know about this line. The equation of the line is y = 4x - 6 The slope is equal to 4 The y-intercept is equal to -6 The x-intercept is equal to 1.5 The line with the steep slope in the following graph is the line that these values are referencing. The second line can be analyzed the same way using the slope-intercept form of the line. We know the slope is 1/8. Slope-Intercept form of the equation for the line is: y = .125*x + b. Since (2,2) are one of the points on this line, we can replace y with 2 and x with 2 in our equation to get: 2 = .125*2 + b We can solve for b to get: b = 1.75 which is the same as (7/4) that we calculated earlier. Alternatively, we can analyze the second line using the point-slope form of the equation for a straight line of (y-y1) = m*(x-x1) by replacing (x1,y1) with (2,2) to get: y-2 = .125*(x-2) We remove parentheses on the right side of this equation to get: y-2 = .125x - .25 We add 2 to both sides of this equation to get: y = .125x - .25 + 2 which becomes: y = .125*x + 1.75 which is the same as: y = (1/8)x + (7/4) The y-intercept is (7/4) which is the same as 1.75. The x-intercept of this line is found by replacing y with 0 and solving for x. y = (1/8)x + (7/4) becomes: 0 = (1/8)x + (7/4). Subtract (7/4) from both sides of this equation to get: -(7/4) = (1/8)x. Multiply both sides of this equation by 8 to get: -14 = x which is the same as x = -14. The equation for this line is : y = (1/8)x + (7/4) which is the same as: y = .125x + 1.75 The slope is equal to .125 The y-intercept is equal to 1.75 The x-intercept is equal to -14. The line with the shallow slope in the following graph is the line referenced by these values. SUMMARY OF FORMS OF EQUATIONS FOR A STRAIGHT LINE There is the standard form of the equation: ax + by = c There is the slope-intercept form of the equation: y = mx + b There is the point-slope form of: (y-y1) = m*(x-x1) CALCULATING THE SLOPE AND Y-INTERCEPT FROM LINEAR EQUATION IN STANDARD FORM You can find the slope and the y-intercept directly from the standard equation if you remember the formula. example: Standard form of equation for a straight line is ax + by + c Slope is (-a/b) and y-intercept is (c/b) Assume your equation is 5x + 6y = 14 Slope is (-5/6) and y-intercept is (14/6) Let's see how we did by converting this equation to slope-intercept form. Subtract 5x from both sides of this equation to get 6y = -5x + 14 Divide both sides of this equation by 6 to get y = (-5/6)*x + (14/6) Slope-Intercept form of this equation is: y = (-5/6)x + (14/6) Slope is (-5/6) and y-intercept is (14/6) Either way you get the same slope and y-intercept. This is no surprise because the slope-intercept form of the equation is derived from the standard form of the equation. If your equation is in standard form of ax = by + c, then you can get slope and y-intercept directly by using the following formulas: Slope = (-a/b) y-Intercept = (c/b) Most of the time, however, you will be converting to slope-intercept form because that gets you the slope and the y-intercept and also puts the equation in a form that can be graphed without further modification. PROPERTIES OF LINES Lines are either parallel to each other, perpendicular to each other, identical to each other, or will intersect with each other when they are not parallel to each other. Parallel lines will never intersect. Perpendicular lines will always intersect at one point only. Identical lines will intersect at all points because they are the same line. Any other lines that are not parallel to each other nor identical to each other will intersect at one point only. This assumes the lines are in the same plane. If you draw 2 lines on one piece of paper that is lying flat, the paper represents the plane that the lines are in. All lines in this lesson are assumed to be in the same plane. PARALLEL LINES Parallel lines are lines that will never intersect no matter how far you extend them in either direction. PARALLEL LINES IN SLOPE-INTERCEPT FORM The slope-intercept form of the line allows you to see if the lines are parallel immediately. The general form of the slope-intercept form of the equation for a line is: y = mx + b m is the slope. b is the y-intercept. If the slopes are equal and the y-intercepts are not, then the lines are parallel. If the slopes are equal and the y-intercepts are also equal, then the lines are identical. The only thing you need to be careful of is that you need to reduce fractions to their simplest form in order to make it clear. Example 1: y = 5x + 6 y = 5x + 6 These lines are identical because they have the same slope and the same y-intercept. Example 2: y = 5x + 6 y = 5x + 3 These lines are parallel because they have the same slope and a different y-intercept. Example 3: y = (1/5)x + 6 y = (2/10)x + 3 These lines are parallel because they have the same slope and a different y-intercept. The slope of (2/10) equals (1/5) after you reduce it to its simplest form. PARALLEL LINES IN STANDARD FORM Standard Form is ax + by = c Parallel lines will have the same a and b terms but different c terms, or: Parallel lines will have a and b terms that are multiples of each other by a common factor but the c terms are not multiples of each other by the same common factor. Example 1: x + y = 3 x + y = 5 These lines are parallel because the a and b terms are the same but the c terms are different. Example 2: x + y = 3 5x + 5y = 12 These lines are parallel because the a and b terms are multiples of each other by a common factor of 5 but the c terms are not. Example 3: x+y = 3 5x + 5y = 15 These lines are identical because the a and b and c terms are multiples of each other by a common factor of 5. To graph these lines, convert them to slope-intercept form. They become: x + y = 3 becomes y = -x + 3 ____________________ (original line from example 1) x + y = 5 becomes y = -x + 5 ____________________ (parallel line from example 1) 5x + 5y = 12 becomes y = -x + (12/5) ____________ (parallel line from example 2) 5x + 5y = 15 becomes y = -x + 3 _________________ (identical line from example 3) Once converted to slope-intercept form, it is very easy to see which lines are parallel because they have the same slope and a different y-intercept. It is also very easy to see which lines are identical because they have the same slope and the same y-intercept. Graph of these lines is shown below: Click on the following hyperlink to see a picture of this graph with comments. Picture of Graph from Equations in Standard Form Examples PARALLEL LINES IN POINT-SLOPE FORM Point-Slope Form is (y-y1) = m(x-x1) Parallel Lines will have the same m term which is the slope of the line. The points (y-y1) and (x-x1) will not be identical. If the x1 terms are identical, then the y1 terms are not. If the y1 terms are identical, then the x1 terms are not. If neither the x1 terms or the y1 terms are identical then it would be very difficult to tell unless the point-slope form is converted to the slope-intercept form. Example 1: (y-1) = 5(x-1) (y-1) = 5(x-2) These lines are parallel because they have the same slope and the y1 terms are identical while the x1 terms are not. Example 2: (y-1) = 5(x-1) (y-3) = 5(x-1) These lines are parallel because they have the same slope and the x1 terms are identical while the y1 terms are not. Example 3: (y-1) = 5(x-1) (y-3) = 5(x-5) These lines are parallel because they have the same slope and the points (1,1) and (5,3) are not on the same line. If the points (1,1) and (5,3)) are on the same line, then these lines would be the same lines and would therefore be identical. It is very difficult to see this without converting the equations to slope-intercept form. Example 4: (y-1) = 5(x-1) (y-6) = 5(x-2) These lines are identical because they have the same slope and the points (1,1) and (2,6) are on the same line. It is very difficult to see this without converting the equations to slope-intercept form. To graph these lines, convert them to slope-intercept form. They become: (y-1) = 5(x-1) becomes y = 5x-4 ______________________ (original line from example 1) (y-1) = 5(x-2) becomes y = 5x-9 ______________________ (parallel line from example 1) (y-3) = 5(x-1) becomes y = 5x-2 ______________________ (parallel line from example 2) (y-3) = 5(x-5) becomes y = 5x-22 _____________________ (parallel line from example 3) (y-6) = 5(x-2) becomes y = 5x-4 ______________________ (identical line from example 4) You can see very easily in slope-intercept form that all these lines except the last one are parallel to the first line because they have the same slope with a different y-intercept. You can also see that the last line is identical to the first line because it has the same slope and the same y-intercept. Graph of these lines is shown below: Click on the following hyperlink to see a picture of this graph with the points that the point-slope form of the equations were built on. Picture of Graph from Equations in Point-Slope Form Examples It can be seen that the original line from example 1 which was generated from point (1,1) is the same line as the identical line from example 4 which was generated from point (2,6). PERPENDICULAR LINES Perpendicular lines are lines that intersect with each other at right angles. PERPENDICULAR LINES IN SLOPE-INTERCEPT FORM The slope-intercept form of the line is y = mx + b m is the slope. b is the y-intercept. Lines are perpendicular to each other if the slopes are negative reciprocals of each other regardless of whether the y-intercepts are equal or not. Negative reciprocal of m = -1/m Example: Line parallel to y = 2x + 2 is y = 2x - 2. Line identical to y = 2x + 2 is y = 2x + 2. To makes these lines perpendicular to y = 2x + 2, simply replace their slope with the negative reciprocal of it. y = 2x - 2 becomes y = -(1/2)x - 2 y = 2x + 2 becomes y = (-1/2)x + 2 Each one of these lines will have the same y-intercept as the line it is perpendicular to because we only changed the slope, not the y-intercept. Graph of these lines is shown below: PERPENDICULAR LINES IN STANDARD FORM Two lines are perpendicular to each other in standard form if: The a term is equal to the b term of the original line, and the b term is equal to the a term of the original line, and the signs of either the a term or the b term are reversed. Example: 5x + 10y = 3 is the original line. -10x + 5y = 3 is a perpendicular line. 10x - 5y is also a perpendicular line. Conversion of all these equations to slope-intercept form in order to be able to graph them yields the following: 5x + 10y = 3 becomes y = -.5x + .3 -10x + 5y = 3 becomes y = 2x + .6 10x - 5y = 3 becomes y = 2x - .6 Graph of these lines is shown below: The original line is slanting down to the right. The two perpendicular lines are slanting up to the right. PERPENDICULAR LINES IN POINT-SLOPE FORM Two lines are perpendicular to each other in point-slope form if the m term of one is a negative reciprocal of the m term of the other. The general form of the point-slope form of the equation for a line is (y-y1) = m(x-x1). m is the slope. Example: Original line is (y-5) = 2(x-3). Line perpendicular to it is (y-5) = -(1/2)*(x-3). Slope-intercept form of the equations for these two lines is: (y-5) = 2(x-3) becomes y = 2x-1. (y-5) = -(1/2)*(x-3) becomes y = -(1/2)x + 3/2 + 5 which becomes y = (-(1/2)x + (13/2) which becomes y = -.5x + 6.5 Graph of these two lines is shown below: In all cases, converting the equation to slope-intercept form is the most reliable way to get the slope and the y-intercept and gives you the bonus of being able to graph the equations without any further modification. In this form: If the lines are parallel, the slopes are equal and the y-intercepts are not. If the lines are identical, the slopes and the y-intercepts are equal. If the lines are perpendicular, the slopes are negative reciprocals of each other. INTERSECTION OF TWO STRAIGHT LINES The intersection of two straight lines is a point that is common to both of those lines. Example: y = 5x + 3 y = -(1/5)x + 3 It just so happens that these lines intersect at the y-intercept. This is because when x = 0 in both equations, y equals 3 in both equations. The point (0,3) is common to both of these equations. A graph of those two equations is shown below. Assuming that we did not know that these lines intersected at the point (0,3), we would find that point by solving both equations simultaneously. SOLVING LINEAR EQUATIONS SIMULTANEOUSLY. When you solve two equations simultaneously, you are looking for a solution that will solve both equations at the same time. Consider the following two equations. y = 3x y = x+2 Both equations have many solutions. x can be any real number and satisfy the first equation and also satisfy the second equation. Solution sets for the first equation would be: (x,y) 1,3 2,6 3,9 4,12 etc. Solution sets for the second equation would be: 1,3 2,4 3,5 4,6 etc. Of all these solution sets, the only solution set that was able to satisfy both equations at the same time was (x,y) = (1,3) We can solve for simultaneous solutions in several ways. SUBSTITUTION METHOD FOR SOLVING SIMULTANEOUS LINEAR EQUATIONS The substitution method takes one of the equations and solves for one of the variables. It then uses the solved value for that variable in the other equation. With two equations in two unknowns, this is a one step process. With three equations in three unknowns, this would be a two step process. We'll do two equations in two unknowns in this lesson and leave the three equations in three unknowns for another lesson. That lesson will be titled LINEAR EQUATIONS IN THREE DIMENSIONS. EXAMPLE 1 You have two equations in two unknowns. They are: 3x + 6y = 21 7x - 8y = 32 Pick one of the equations and solve for y or x. It doesn't matter which one so pick any one that you want. We'll try the first equation and solve for x Equation to solve for x is: 3x + 6y = 21 Subtract 6y from both sides to get: 3x = 21-6y Divide both sides by 3 to get: x = (21-6y)/3 = 7-2y We have: x = 7-2y Use this solved variable in the second equation. The purpose of this is to transform the second equation into one equation in one unknown which is solvable. Once we solve for that unknown we can use the value obtained for that unknown to solve for the other unknown. Our second equation is: 7x - 8y = 32 Replace x with 7-2y to get: 7*(7-2y) - 8y = 32 which becomes: 49 - 14y - 8y = 32 Combine like terms to get: 49 - 22y = 32 Subtract 49 from both sides to get: -22y = 32-49 = -17 We have: -22y = -17 Divide both sides by -22 to get: y = 17/22 We have solved for the value of one of the unknowns. Go back to the first equation and solve it after replacing y with 17/22. First equation is: 3x + 6y = 21 Replace y with 17/22 to get: 3x + 6*(17/22) = 21 Simplify to get: 3x + (6*17)/22 = 21 which equals 3x + 102/22 = 21 which becomes 3x + 51/11 = 21. Multiply both sides of this equation by 11 to get: 33x + 51 = 231 Subtract 51 from both sides of this equation to get: 33x = 231 - 51 = 180 Divide both sides of this equation by 33 to get: x = 180/33 The solution to this set of simultaneous equation is: x = 180/33 y = 17/22 This solution must satisfy both equations at the same time. to confirm this solution is good, we substitute 180/33 for x and 17/22 for y in the original equations to confirm that they are true. Equations are: 3x + 6y = 21 7x - 8y = 32 First equation is 3x + 6y = 21 Substituting in the first equation gets: 3*(180/33) + 6*(17/22) = 21 Simplify to get: 180/11 + 51/11 = 21 Simplify further to get: (180+51)/11 = 21 which becomes 231/11 = 21 which becomes: 21 = 21 confirming the values for x and y are good in the first equation. Second equation is 7x - 8y = 32 Substituting in the second equation gets: 7*(180/33) - 8*(17/22) = 32 Simplify to get: (1260/33) - (136/22) = 32 Simplify further to get: (420/11) - (68/11) = 32 Simplify further to get: (420-68)/11 = 32 which becomes 352/11 = 32 which becomes: 32 = 32 confirming the values for x and y are good in the second equation. Since the values of x and y satisfy both equations at the same time, they have solved the equations simultaneously. ELIMINATION METHOD FOR SOLVING SIMULTANEOUS LINEAR EQUATIONS The elimination method uses any combination of multiplying or dividing both sides of an equation by a factor in order to modify one or both of the equations so that, when you add or subtract the two equations together, one of the unknown variables disappears. Sometimes you have to multiply both equations to get a common coefficient that will allow the elimination. Other times you only have to operate on one of the equations. EXAMPLE 1 For the first example, we'll do the same equation we solved using the substitution method. This time, however, we'll use the elimination method. You have two equations in two unknowns. They are: 3x + 6y = 21 7x - 8y = 32 You want to eliminate either the x variable or the y variable. It doesn't matter which one as long as one of them goes away. Looking at these 2 equations, it does not appear that there is a simple way to do it. 7 is not a multiple of 3 directly. 6 is not a multiple of 8 directly. we can, however, multiply both equations to get a common coefficient in one of the variables so that it can be eliminated. We'll do the x variable. 3 * 7 = 21 7 * 3 = 21 we'll be able to get a common coefficient for x that will then allow us to eliminate that variable. We multiply both sides of the first equation by 7 and we multiply both sides of the second equation by 3. They become: 21x + 42y = 147 (first equation was multiplied by 7) 21x - 24y = 96 (second equation was multiplied by 3) We now have a common coefficient for the x term. If we subtract one equation from the other, the x variable will cancel out. We choose to subtract the second equation from the first to get: 66y = 51 Do not forget that when you subtract a minus from a plus, it's the same thing as adding a plus to a plus. 42 - (-24) = 42 + 24 = 66 Always be careful of the signs when you are adding or subtracting or multiplying or dividing. I only tell you this because I have been caught making mistakes too many times for me to even want to remember. A simple error like this can destroy even the most sophisticated solutions to a problem. We divide both sides of this equation by 66 to get: y = 51/66 We simplify this to get: y = 17/22 Since this is the same answer for y we got earlier using the substitution method, we're on the right track. Now that we have a value for y, we can solve for x. We take either equation and replace y with 17/22 and solve. Let's take the second equation of: 7x - 8y = 32 Replace y with (17/22) to get: 7x - 8*(17/22) = 32 Simplify to get: 7x - 136/22 = 32 Multiply both sides of this equation by 22 to get: 154x - 136 = 704 Add 136 to both sides to get: 154x = 840 Divide both sides by 154 to get: x = 840/154 Simplify to get: x = 180/33 Please note that x = 180/33 is accurate but that it was not easy to see that 840/154 was the same as 180/33. The common factor appears to be to multiply both numerator and denominator by (3/14). 840*(3/14) = 180. 154*(3/14) = 33. We got the same result using the elimination method as using the substitution method. You can choose to use either one, whichever is easier for you. They both work equally well. GRAPHING METHOD FOR SOLVING SIMULTANEOUS LINEAR EQUATIONS The graphing method plots both equations on a graph and then shows where the intersection of the two lines is. That intersecting point is the common solution to both equations. In order to graph the equations, you have to convert them to the slope-intercept form of a straight line. EXAMPLE 1 We'll use the same equation we solved using the substitution and elimination method. This time, however, we'll use the graphing method. You have two equations in two unknowns. They are: 3x + 6y = 21 7x - 8y = 32 In order to solve them using the graphing method, you have to convert them to the slope-intercept form of the equation if they are not already in that form. First equation is: 3x + 6y = 21 Subtract 3x from both sides to get: 6y = -3x + 21 Divide both sides by 6 to get: y = (-3/6)x + (21/6) Second equation is: 7x - 8y = 32 Subtract 7x from both sides to get: -8y = -7x + 32 Divide both sides by -8 to get: y = (7/8)x + (32/-8) which becomes: y = (7/8)x - (32/8) Your two equations are now: y = (-3/6)x + (21/6) y = (7/8)x - (32/8) Graph of these equations looks like the following: We have an intersection at somewhere around x = 5.5 and y = .78 Our known solution of y = 17/22 is equivalent to .772727273 which is close to .78. Our known solut9ion of x = 180/33 is equivalent to 5.454545455 which is close to 5.5. The graph has brought us close to our solution, but not exactly on it. That is the main difficulty in solving by graphing. It's difficult to pinpoint the exact solution. You can get close, but rarely will you be right on. The graphing method to solve simultaneous equations is not recommended if you need to pinpoint your solution. It is recommended as a sanity check to see if the solution you have solved in other ways is confirmed by the graph. It is also recommended as a quick way to see if you have any solutions at all, and, if so, what general area those solutions would be in. Some graphing programs will show you the (x,y) coordinates of any point you select. If you move your mouse over the point, the software will tell you the coordinates of your mouse. This is more accurate but still open to error. The most accurate is to solve the equation using one of the other methods. The following hyperlinks shows an example of a graphing calculator that was able to zoom in on, and show the coordinates of, the solution. Click on the following hyperlink to see: Graphing Solution for Simultaneous Equations Demonstration WHEN THE SOLUTION TO THE SOLVING OF SIMULTANEOUS EQUATIONS IS AN IDENTITY Consider the following two equations. 5x + 6y = 36 10x + 12y = 72 Solve these two equations using the elimination method. Multiply the first equation by 2 to get: 10x + 12y = 72 Your two equations that have to be solved simultaneously become: 10x + 12y = 72 10x + 12y = 72 Subtract second equation from the first to get: 0 + 0 = 0 This is called an identity solution. These equations are identical. If you solve for one of these equations, you have solved for both since they represent the same line. These lines are identical and they led to a true equation even though both variables disappeared in the solution. 0 = 0 is a true equation. What if the lines were parallel? Consider the following equations: 5x + 6y = 30 10x + 12y = 72 Multiply the first equation by 2 to get: 10x + 12y = 60 10x + 12y = 72 Subtract the first equation from the second to get: 0 + 0 = 12 resulting in: 0 = 12 Both variables disappeared, only this time the equation is not true (it is false). This means there is no simultaneous solution to these equations. The following graph of these two equations will show that they are parallel and will never intersect. You are encouraged to look at the references. There are numerous examples and problems in there for you to see and test your knowledge. Questions or comments regarding this lesson can be directed to dtheophilis@yahoo.com. This lesson has been accessed 220 times. |
