Machinery's Handbook, 31st Edition
32 Equation Solving The first equation is rearranged to y = 7 x – 3. Substituting into the second equation yields: 14 2 7 3 6 x x − − = ( ) → − + = → = 14 14 6 6 6 6 x x . If a system results in a = a , then the lines are the same, or collinear. All of the points on either line are solutions to (points on) the other line. The indication that lines are collinear is that one is a multiple of the other, term by term. Example (No Solution): Parallel lines have the same slope, and so they have no point of intersection. This is seen in the substitution of one equation’s variable into the other equation, giving a false statement, such as 1 = 3. An example is the system y = – x + 4 and y = – x – 1. Substitution of the first into the second gives – x + 4 = – x – 1, or 4 = –1. This clearly is not true, hence the system has no solution. A final way to find the solution to a linear system is to put the coefficients of the equations into the following formulas and to solve for x and y , as follows:
a 1 x + b 1 y = c 1 a 2 x + b 2 y = c 2
Then,
c 1 b 2 c 2 b 1 – a 1 b 2 a 2 b 1 – = ---------------
a 1 c 2 a 2 c 1 – a 1 b 2 a 2 b 1 – = ---------------
x
y
Example:
3 x 4 y + = 17 5 x 2 y – = 11
17 ( ) –2 ( ) 3 ( ) –2 ( ) 5 ( ) – 4 ( ) ------------------------ 11 ( ) – 4 ( )
– 34– 44 – 6– 20 -----------
–78 –26 ----- 3
= =
=
=
x
The value of y can now be most easily found by inserting the value of x in one of the equations: 5 ( ) 3 ( ) 2 y – = 11, 2 y = 15– 11 = 4, y = 2 Checking the solution by putting these values into the original system shows that (3, 2) is the solution of this linear system. Solving a Second-Degree (Quadratic) Equation.— A second-degree equation is also called a quadratic equation . To solve a second-degree equation is to find the x values (or points) at which the parabola intersects the x -axis. These are the roots or zeros of the parabola. There are several ways to solve a quadratic equation for its roots. If an equation can be factored, then each factor is set equal to zero and the solution is thus found. This is according to the zero property of multiplication , which states that if AB = 0, then either A = 0 or B = 0. If the equation is not readily or obviously factored (or even if it is), the quadratic for- mula can be used to find the roots, as explained in the next section, Using the Quadratic Formula . The simplest quadratic equation has the form x 2 = c , where c is a constant. Solving this entails simply taking the square root of both sides: x 2 = c , therefore, x = c and – c . For example, x 2 = 36 has two solutions, x = 6 and x = –6. If the quadratic has the form, like ax 2 = c , the solution is also straightforward, x 2 = a / c , and so x = a c ± / . Note: The square root of a number as it stands alone is understood to be its positive root; that is, the square root of 9 is 3. But in an equation the solution includes both positive and negative roots. This makes sense because the equation represents a parabola, which can intersect the x -axis in two places, its two roots. The other two possibilities occur when the parabola (a quadratic function) intersects the x -axis once (touches it) or does not inter- sect it. All three possibilities are revealed in solutions to the quadratic equation. Designating the roots of a quadratic equation (or any polynomial) by r , x 2 = c can be regarded as x 2 = r 2 , so x = r and – r , the roots of the parabola. Another way to rewrite x 2 = r 2
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