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Solution for -9x+1=-x+17 equation:Simplifying -9x + 1 = -1x + 17 Reorder the terms: 1 + -9x = -1x + 17 Reorder the terms: 1 + -9x = 17 + -1x Solving 1 + -9x = 17 + -1x Solving for variable 'x'. Move all terms containing x to the left, all other terms to the right. Add 'x' to each side of the equation. 1 + -9x + x = 17 + -1x + x Combine like terms: -9x + x = -8x 1 + -8x = 17 + -1x + x Combine like terms: -1x + x = 0 1 + -8x = 17 + 0 1 + -8x = 17 Add '-1' to each side of the equation. 1 + -1 + -8x = 17 + -1 Combine like terms: 1 + -1 = 0 0 + -8x = 17 + -1 -8x = 17 + -1 Combine like terms: 17 + -1 = 16 -8x = 16 Divide each side by '-8'. x = -2 Simplifying x = -2 You can always share this solutionSee similar equations:| (x^2+4)x(x^3-27i)=0 | | x+2=5x-10 | | 4(9-7x)=0 | | (logx+4)-logx=log(x+1) | | 9x+1=-x+17 | | 9-7x=0 | | 5x+35-4x-36=4x-13 | | 2x^3-3x^2+10x-15=0 | | 2x+24=2x^2 | | 3(a+6)-4=12a+2(10+4a) | | 2x-3=-7x+5 | | -1-3p=2 | | -2(5x-3)15x=-8x+18x+51 | | 1.3b+3.7b=9 | | 3x-1=59 | | tan(6x)=1 | | a-3a=12-6 | | -4a+9b+2b-11b= | | -r^2-4r+45=0 | | g+38.75=41.5 | | -y^3+2y^2+8=0 | | 4b+6-2=6+4 | | 40+9x=148 | | 3s+2=17 | | LN(4x)=3.5 | | 12/15x=9/135 | | 5(2c+3)=35 | | 6=2(3+4)+(n-7) | | -4+12x=-7 | | 21+5x=2x-33 | | 20z=x^3-x^2 | | 3m-9=3m+9 | Algebra ExamplesPopular Problems Algebra Solve for x -9x+1=-x+17 Step 1 Move all terms containing to the left side of the equation. Tap for more steps... Step 1.1 Add to both sides of the equation. Step 1.2 Add and . Step 2 Move all terms not containing to the right side of the equation. Tap for more steps... Step 2.1 Subtract from both sides of the equation. Step 2.2 Subtract from . Step 3 Divide each term in by and simplify. Tap for more steps... Step 3.1 Divide each term in by . Step 3.2 Simplify the left side. Tap for more steps... Step 3.2.1 Cancel the common factor of . Tap for more steps... Step 3.2.1.1 Cancel the common factor. Step 3.2.1.2 Divide by . Step 3.3 Simplify the right side. Tap for more steps... Step 3.3.1 Divide by . 7\left(7+2x\right)=3\left(9x-1\right) Multiply both sides of the equation by 21, the least common multiple of 3,7. 49+14x=3\left(9x-1\right) Use the distributive property to multiply 7 by 7+2x. 49+14x=27x-3 Use the distributive property to multiply 3 by 9x-1. 49+14x-27x=-3 Subtract 27x from both sides. 49-13x=-3 Combine 14x and -27x to get -13x. -13x=-3-49 Subtract 49 from both sides. -13x=-52 Subtract 49 from -3 to get -52. x=\frac{-52}{-13} Divide both sides by -13. x=4 Divide -52 by -13 to get 4. Contents: This page corresponds to § 2.4 (p. 200) of the text. Suggested Problems from Text: p. 212 #7, 8, 11, 15, 17, 18, 23, 26, 35, 38, 41, 43, 46, 47, 51, 54, 57, 60, 63, 66, 71, 72, 75, 76, 81, 87, 88, 95, 97
Quadratic EquationsA quadratic equation is one of the form ax2 + bx + c = 0, where a, b, and c are numbers, and a is not equal to 0. FactoringThis approach to solving equations is based on the fact that if the product of two quantities is zero, then at least one of the quantities must be zero. In other words, if a*b = 0, then either a = 0, or b = 0, or both. For more on factoring polynomials, see the review section P.3 (p.26) of the text. Example 1.
Square Root PrincipleIf x2 = k, then x = ± sqrt(k). Example 2.
Example 3. Example 4.
Completing the SquareThe idea behind completing the square is to rewrite the equation in a form that allows us to apply the square root principle. Example 5.
Example 6.
So far we have discussed three techniques for solving quadratic equations. Which is best? That depends on the problem and your personal preference. An equation that is in the right form to apply the square root principle may be rearranged and solved by factoring as we see in the next example. Example 7. x2 = 16. x2 - 16 = 0. (x + 4)(x - 4) = 0. x = -4, or x = 4. In some cases the equation can be solved by factoring, but the factorization is not obvious. The method of completing the square will always work, even if the solutions are complex numbers, in which case we will take the square root of a negative number. Furthermore, the steps necessary to complete the square are always the same, so they can be applied to the general quadratic equation ax2 + bx + c = 0. The result of completing the square on this general equation is a formula for the solutions of the equation called the Quadratic Formula. Quadratic Formula
We are saying that completing the square always works, and we have completed the square in the general case, where we have a,b, and c instead of numbers. So, to find the solutions for any quadratic equation, we write it in the standard form to find the values of a, b, and c, then substitute these values into the Quadratic Formula. One consequence is that you never have to complete the square to find the solutions for a quadratic equation. However, the process of completing the square is important for other reasons, so you still need to know how to do it! Examples using the Quadratic Formula: Example 8.
Example 9.
Example 10.
The expression under the radical in the Quadratic Formula, b2 - 4ac, is called the discriminant of the equation. The last three examples illustrate the three possibilities for quadratic equations.
Notes on checking solutions
Exercise 1:
Return to Contents Equations Involving RadicalsEquations with radicals can often be simplified by raising to the appropriate power, squaring if the radical is a square root, cubing for a cube root, etc. This operation can introduce extraneous roots, so all solutions must be checked. If there is only one radical in the equation, then before raising to a power, you should arrange to have the radical term by itself on one side of the equation. Example 11.
Look at what would have happened if we had squared both sides of the equation before isolating the radical term. This is worse than what we started with! If there is more than one radical term in the equation, then in general, we cannot eliminate all radicals by raising to a power one time. However, we can decrease the number of radical terms by raising to a power. If the equation involves more than one radical term, then we still want to isolate one radical on one side and raise to a power. Then we repeat that process. Example 12.
Note on checking solutions:
Exercise 2:
Return to Contents Polynomial Equations of Higher DegreeWe have seen that any degree two polynomial equation (quadratic equation) in one variable can be solved with the Quadratic Formula. Polynomial equations of degree greater than two are more complicated. When we encounter such a problem, then either the polynomial is of a special form which allows us to factor it, or we must approximate the solutions with a graphing utility. Zero ConstantOne common special case is where there is no constant term. In this case we may factor out one or more powers of x to begin the problem. Example 13.
Factor by GroupingExample 14.
Quadratic in FormExample 15.
Exercise 3:
Return to Contents Equations Involving Fractional Expressions or Absolute ValuesExample 16.
Example 17.
Exercise 4:
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