 Solving Algebraically SUBSTITUTION Example 1: y = 2x + 3 4x + 3y - 19 = 0 If there is a solution point, the two equations will both share that point. We're going to mix the two equations by substituting one equation into the other. Specifically, we're going to substitute the 2x + 3 value for y from the first equaton into the second equation where there is a y: 4x + 3(2x + 3) - 19 = 0 and now we'll simplify: 4x + 3(2x + 3) - 19 = 0 4x + 6x + 9 - 19 = 0 10x -10 = 0 We can solve this for x: 10x = 10 x = 1 This represents the x coordinate of the solution point. We can substitute it into either equation to find the corresponding y value: (we'll use the first equation because it's simpler) y = 2x + 3 = 2(1) + 3 = 5 So the solution point is (1, 5) You may be asked to verify this solution by substituting it into each of the equations to show that it makes each true.
Important Note: When substituting the point to check it, always use the ORIGINAL two equations. If you'd made an algebra mistake in later lines, the point you find might check in those but still be incorrect! The solution is (1, 5). It is the only point which will work in both equations. [We won't show these checks in future examples] Example 2: x + 4y = -5 5x - 3y - 21 = 0 The first equation can be rearranged slightly to solve this time for x: x = -4y - 5 Now we can use this expression -4y - 5 to substitute into the second equation for x: 5(-4y - 5) - 3y - 21 = 0 and simplify: -20y - 25 - 3y - 21 = 0 -23y - 46 = 0 now solve for y: -23y = 46 y = -2 This represents the y coordinate of the solution point. We can substitute it into either equation to find the corresponding x value: We'll use the first equation, as it's less complicated. Important Note: When substituting back in to find the other value, always use one of the ORIGINAL two equations. If you'd made an algebra mistake in a later line, the point you find might be incorrect but still check! Substitute y = -2 into the first equation: x + 4(-2) = -5 x - 8 = -5 x = 3 So the solution point is (3, -2) [In principle, you could rearrange either equation to solve for x or y, and substitute. In practice, however, this will be too much work, especially as fractional coefficients often occur, making the steps more difficult. There is an easier method (elimination) we'll use instead later.] Example 3: y = -2x + 1 6x + 3y - 4 = 0 Substitute -2x + 1 for y in the second equation: 6x + 3(-2x + 1) - 4 = 0 6x -6x + 3 - 4 = 0 -1 = 0 The x's disappeared, and we're left with a statement that isn't true. When we mixed the equations together, we assumed there was a solution. This result tells us that the equations don't mix; there is no intersection point because the lines are parallel. There is no solution Example 4: x = ½y + 5 6x - 3y - 30 = 0 Substitute ½y + 5 for x in the second equation: 6(½y + 5) - 3y - 30 = 0 3y + 30 - 3y - 30 = 0 and solve for y: 0 = 0 The y's disappeared, and we're left with a statement that is always true. When we mixed the equations together, we assumed there was one unique solution. This result tells us that the equations mix everywhere; there is no single intersection point because the lines are the same. There is no solution Shortcuts for Standard Form Equations:
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