Algebra Know-It-ALL
There are only two integer factors n of the leading coefficient: 1 and −1. All the possible ratios m/n are the same as the inte ...
446 Polynomial Equations in Real Numbers Using synthetic division, find a lower bound and an upper bound for the interval conta ...
We’ve seen how pairs of linear equations can be solved as two-by-two systems. What if the equations are more complicated? In thi ...
If we call x the independent variable, then the quadratic equation is already expressed as a function of x, so we don’t have to ...
For x= 3, we have y= 2 × 3 + 1 = 6 + 1 = 7 The two solutions of the system are therefore (x, y)= (−2,−3) and (x, y)= (3, 7). Fin ...
Are you confused? We’ve found two ordered pairs (x,y) that solve the above pair of equations as a two-by-two system. How do we k ...
Now we have a quadratic equation in standard form. That means we can solve it directly with the quadratic formula. Let’s state t ...
and 3 x^2 +y+ 5 x− 11 = 0 First, we morph In both of these equations, a multiple of y can be separated out and placed alone on t ...
and x− 3 = 0 giving us x=−5 or x= 3. We can substitute these two values for x into either of the original functions to get the y ...
4 × 9 + 18 + (−62)+ 8 = 0 36 + 18 + (−62)+ 8 = 0 54 + (−62)+ 8 = 0 − 8 + 8 = 0 0 = 0 Next, we check (−5,−39) in the second origi ...
Solution Both of these equations are presented as functions of x, so we can mix the right sides directly without hav- ing to man ...
the denominator. But there’s no good reason to expand the products of the binomials. That would produce a formula with fewer gro ...
We can add y to each side and transpose the equation left-to-right, getting the function y= 3 x+ 1 Next, we mix When we mix the ...
Our second solution is (x,y)= (−2,−5). Plugging in x=−3, we have y= 3 x+ 1 = 3 × (−3)+ 1 =− 9 + 1 =− 8 Our third solution is (x, ...
Next, we check (−2,−5) in the second original equation: − 6 x+ 2 y= 2 − 6 × (−2)+ 2 × (−5)= 2 12 + (−10)= 2 2 = 2 Finishing up, ...
Now here’s the trick! If we spend enough time playing around with this, we’ll find that it can be factored into (3x+ 2)(x^2 + 1) ...
=j 2 + (−1)+ (−j2)+ 5 =j 2 +−j 2 + (−1)+ 5 = 0 + 4 = 4 Our third solution is (x,y) = (−j, 4). And finally ... Now it’s time to c ...
Solve the following pair of equations as a two-by-two system, including the complex- number solutions, if any. Let x be the ind ...
In this chapter, we’ll graph the systems we solved in Chap. 27 to see how they work in the realm of real numbers. Real solutions ...
Table 28-1 compares some values of x, some values of the first function, and some values of the second function. The left-hand c ...
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