Basic Equation Solving Strategies

Basic Equation Solving Strategies Case 1: The variable appears only once in the equation. (Use work backwards method.) 1 1. Simplify both sides of the equation if possible.. Apply the order of operations rules to make an equation solve plan.. Starting with the last step in the solve plan, apply the same operation to both sides of the equation that reverses the step in the solve plan. Continue working backwards towards the variable being solved for by reversing or undoing each step in the solve plan.. Check the solution. Substitute each solution into the original equation to see if the resulting equation is true. Basic Operation and the Corresponding Reverse Operation a) + <---------> - (Addition and subtraction reverse each other.) b) * <----------> / (Multiplication and division reverse each other.) c) x <----------> or - (Raising to a power and root operation reverse each other.) x <-----------> Example 1: If w = 11, then w = 11 or w = -11. Example : If (y ) = 89, then y - = 17 or y - = -17 y = 0 or y = -1. Example : If (y -) = -89, then (y ) can NOT be reversed! Why? Example : If (x ) = 15, then x - = 5 ====> x = 7 Example 5: If (x ) = -15, then x - = -5 ====> x = - d) x <----------> x or -x (x and -x have the same absolute value.) Example 1: If t = 1, then t = 1 or t = -1. Example : If w = 0, then w = 0 or w = -0 Example : If w = -0, then w can NOT be reversed!! Why? e) -x <---------> -x (Taking the opposite of a number is its own reverse operation.) 1/x <--------> 1/x (Taking the reciprocal of a number is its own reverse operation.) Ex. 1 0x + 50 = 770 Ex. x - 17 = -0 Ex. y - 17 = -0 10 10 x 0x = 70 (- 50) x x - 17 = -00 (* 10) x y = - (+ 17) 10 x = (/ 0) * x = -18 (+ 17 ) * *0 y = 0 ( * 10) -17 x = - 5.75 (/ ) / 10 + 50 / 10-17 y = -7.5 ( / )

Ex. 5t + 9-10 = - Ex. 5 (y -7) + 1 = 90 Ex. 6 w = 60 7 t y (y -7) = 89 ( -1) w 5t + 9 = 6 (+10) w = 0 ( * 7) * 5 * * y 7 = 17 or y 7 = -17 w = 0 or -0 + 9 5t + 9 = 18 (*) - 7 ± - w = or -17 ( + ) x 5t + 9 = (x ) x y = or y = -10 ( + 7) w = 105.75 or -10.5 / 5t = 15 (-9) + 1 y = 1 or y = -5 ( / ) / 7-10 t = 6 (/5) ( Undo ) Case : The variable appears more than once in the equation and is raised to the same power through out the equation. 1. Simplify both sides of the equation if possible. Use the distributive property to remove parentheses and combine like terms. At this stage you are only simplifying each side of the equation separately.. Gather the variable on the same side of the equation by adding or subtracting a multiple of the variable to both sides of the equation. ( Note: x, x, x, x, 5x, 6x,... are multiples of x. ). The variable should now appear only once in the equation. Use the work backwards method to solve the equation.. Check your solution. Substitute each solution into the original equation to see if the resulting equation is true. Ex. 1 5(x ) - x + 1 = (x + 10) - 8 Ex. 6x + 18 = -(x 8) - 5x 0 -x + 1 = x + 0 8 (Distributive prop.) x + 6 = -x + - ( Dist. prop. to simplify) x 8 = x + 1 (Simplify both sides of equation.) x + 6 = -x + 0 (Simplify right side.) x x 8 = 1 (Gather variable, subtract x) x 5x + 6 = 0 (Gather variable, add x) * x = 0 (Add 8) * 5 5x = 1 (Subtract 6) - 8 x = 10 (Divide by ) + 6 x =.8 (Divide by 5) Ex. 5x - (10 x) + 1 = x (x - 0) Ex. y + y + y + y - 10 = 186 5x -10 + x + 1 = x x + 0 (Remove ( ) ) y y 10 = 186 (Simplify left side of the equation.) 7x + = x + 0 (Simplify both sides) x y = 196 x 5x + = 0 (Gather variable, subtract x) * y = 9 * 5 5x = 18-10 y = 7 or y = -7 + x = 18/5 =.6

Case : The equation is a simple proportion. (Use the cross product property of fractions.) 1. Simplify both sides of the equation if possible.. Use the cross product property of fractions or proportions to rewrite the equation in equivalent form.. Solve the resulting equation by using the methods discussed above.. Check your solution. Substitute each solution into the original equation to see if the resulting equation is true. ( Think over 1 ) Ex 1: y = Ex. : = Ex. : w = Ex. : x = y + 6 7 x - 5 x 18 w x - 7y 1 = y + x = x 0 w = 6 x = x - 8 y 1 = -x = -0 w = 6 or -6 -x = -8 y = 5 x = 0 x = y = 15 Case : The equation involves different powers of the variable. (Try the factoring method.) 1. Set one side of the equation equal to zero by adding or subtracting algebraic expressions to both sides of the equation.. If necessary, simplify and rewrite the equation so that it is easier to factor.. Factor the equation.. Set each factor that contains a variable to zero and solve each of the resulting equations. 5. Each solution obtained in step () above should be a solution to the original equation. 6. Check all solutions with the original equation. This will help you to better understand what you have found and develop the basic skill of evaluating expressions. Ex. 1 16t = 6t Ex. x - 6x = 5 Ex. x = 8x + x 16t 6t = 0 x 6x 5 = 0 x - 8x - x = 0 (Factor out common factor of 16t ). 16t (t ) = 0 (x x 15) = 0 x(x x 1) = 0 ( Factor out common factor of.) 16t = 0 or t = 0 (x 5)(x + ) = 0 x(x 7)(x + ) = 0 (Use reverse FOIL to factor trinomial. ). t = 0 or t = x 5 = 0 or x + = 0 x = 0 or x 7 = 0 or x + = 0 (Factor out common factor of x). (Use reverse FOIL to factor trinomial. ). t = 0 or t = x = 5 or x = - x = 0 or x = 7 or x = - Ex. y = y Ex. 5 x (6x + 7) = x Ex. 6. z 1z + 6 = 0 y 6 5 ( Largest exponent is twice next largest.) ( Think x over 1 ) y 6y = y 16 6x + 7x = 5x (z )(z 9) = 0 (Use reverse FOIL to factor trinomial. ). y -10y + 16 = 0 6x + 7x 5x = 0 (z + )(z -)(z + )(z ) = 0 ( Factor as difference of two squares. ). (y 8)(y ) = 0 x(6x + 7x 5) = 0 z = - or z = or z = - or z = ( Factor out common factor of x.) y = 8 or y = x(6x 5)(x + 7) = 0 (Use reverse FOIL to factor trinomial. ). x = 0 or 6x 5 = 0 or x + 7 = 0 x = 0 or x = 5/6 or x = -7

Case 5: The equation involves fractions, but the equation is not a proportion. 1. Simplify each side of the equation if possible.. Find the lowest common denominator of all of the fractions that appear in the equation.. Multiple both sides of the equation by the lowest common denominator. This will produce an equation with no fractions!. Simplify each side of the equation by using the distributive property and then combine like terms. 5. Solve the resulting equation by using the methods discussed above. 6. Check all solutions with the original equation. Step () may have created solutions that do not satisfy the original equation. 1 1 5 Ex. 1: + = Ex. : + = 10 Ex. 5 x x + x + = x -1 x -1 æ 1 ö 1 15xç + = 15x è 5 ø x 5x+ 6 x = 15 11x = 15 x = 15 /11 = 1 11 æ 5 ö ( x + ) ç + = ( x + )10 è ( x + ) ø ( x + ) + 15 = 0( x + ) x + 8 + 15 = 0x + 60 x + = 0x + 60 = 6x + 60-7 = 6x -7 x = = -1.0769 6 æ x + ö ( x - 1) ç = ( x -1) è x -1 ø ( x -1) x + = x = 1 Substituting 1 for x in the original equation results in division by zero! = 0 0 Since the only possible solution is not a solution, there are NO solutions!!! Ex. y -50 + y = 5 5y - 5 y - 50 + y = 5 5( y - 5) æ ö (y - 50) 5( y - 5) ç + y = 5( y - 5) è 5 ø 5( y - 5) ( y - 5) + 5 y( y - 5) = y - 50 y - 0 + 5y - 5y = y - 50 5y - 1y - 0 = y - 50 5y - 5y + 0 = 0 5( y - 5y + 6) = 0 5( y - )( y - ) = 0 y - = 0 or y - = 0 y = or y = ( Factor the denominator on the right side of equation. It is easier to find the lowest common denominator when all denominators are factored.) ( Multiply both sides of the equation by 5(y-5) which is the lowest common denominator of all of the fractions in the equation.) ( Simplify both sides of the equation. We have no more factions to deal with!!) ( Use the distributive property to remove grouping symbols.) ( Simplify the left side of the equation. We are dealing with a case situation.) ( Set one side of the equation to zero.) ( Factor out the common factor of 5.) ( Use reverse FOIL to factor the trinomial y 5y + 6 ) ( Set each factor that contains a variable equal to zero.) ( Solve resulting equations. Both solutions check with the original equation.)

Case 6: The equation is a nd degree polynomial equation. (Use the quadratic formula to solve the equation.) Quadratic formula: If ax + bx + c = 0, then x = -b + b - ac a Ex. 1 x x = 15 Ex. -16t + 8t + 0 = 100 x x 15 = 0 ( Set the right side of the equation to 0.) -16t + 8t - 80 = 0 ( Set the right side of the equation to 0.) ( Determine the values a = 1, b = -, and c = -15 a = -16, b = 8, and c = -80 ( Determine the values of a, b and c. ) of a, b and c. ) x = + -(-15) into the quadratic formula.) x = -8 + 7,056 - (1,80) - x = + 6 or - 6 x = -8 + 196 or x = -8-196 to evaluate each part of the - - x = 5 or x = - -1 x = 8-196 or x = 8 + 196 (Multiply by ) -1 Comment: It would have been much easier to use the factoring method to solve this equation. If you can quickly factor x = 1.5 or x = ( Multiply the numerator and denominator the polynomial, then use the factoring method. Otherwise just bite of each fraction by -1 which will reduce the bullet and use the formula. The quadratic formula always works! the number of negative signs. ) Ex.. Solve y + 1y = 9 Ex. Solve x / - 7x 1/ + 1 = 0 At first glance it would appear that the quadratic formula could not be used on this equation. However, if the value of the largest exponent is ( Plug the values of a, b, and c ( Use the order of operation rules formula. ) The strategy used in example will be used to solve this equation. is twice the value of the lower variable exponent, x / - 7x 1/ + 1 = 0 the equation can be rewritten so that the quadratic formula applies. (x 1/ ) -7(x 1/ ) 1 + 1 = 0 5 ( Plug the values of a, b, and c into the quadratic formula.) y + 1y = 9 a = 1, b = -7, and c = 1 ( Determine the values of a, b and c. ) (y ) + 1(y ) 1-9 = 0 x 1/ = 7 + 9 -(1) ( Determine the values a = 1, b = 1, and c = - 9 of a, b and c. ) y = -1 + 1 (-9) x 1/ = 7 + 1 or 7 1 y = -1 + 180 or -1-180 x 1/ = 7 + 1 or 7-1 y = 0.70809 or y = -1.708 x 1/ = or y = 0.70809 ====> y = 0.815 or y = -0.815 y = -1.708 =====> y =.569 i or y = -.569 i. x 1/ = ====> x = 7 x 1/ = ====> x = 6 ( 1/ power = cube root. So cube both sides. ) Comment: If the situation dictates that real number solutions are the only solutions of interest, then ignore the two complex number solutions. i = -1 and i = -1

The Discriminant 6 Ex. 5 Solve x - 6x + = 0 The quantity b ac used in the quadratic formula is called the discriminant. The discriminant reveals the nature a =, b = -6, and c = of the roots and some properties of the parabolic graph. x = 6 + 6 -(1) 1. If b ac > 0 (positive), the polynomial equation will have 8 two different real roots and the graph of the parabola will x = 6 + -1 i = cross the x-axis at two different points. -1 and i = -1 ( i)( i) = 1i = -1 8. If b ac = 0, the polynomial equation will have exactly one real number root and the vertex of the parabola will x = 6 + -1 or 6 - -1 (1 = * ) just touch the x-axis at one point. 8 8. If b ac < 0 (negative), the polynomial equation will have x = 6 + i or 6 - i no real roots, but two complex number roots. The parabola will 8 8 not cross the x-axis. x = + i or - i. In examples 1 of this section the discriminant was positive and consequently each of these equations had two real roots. In example 5 the discriminant was negative and the equation had two complex roots. Comment: The two solutions of this equation are complex numbers which are conjugates of each other. Notice that the solutions are written in standard a + bi format. The graph of y = x - 6x + does not cross the x-axis since the equation x -6x + = 0 has no real number solutions. Since the leading coefficient of the equation is positive, the parabola opens upward and therefore the graph of the parabola is above the x-axis. Refer to the properties of the discriminant above. The x-coordinate of the minimum point of the parabola equals ¾ which equals -b/(a). Solve for a Variable in an Equation (Rearrange the Equation) In each example below, one of the basic equation solving strategies is used to solve for the indicated variable. Ex. 1 1x 6y = 18 (for y) Ex. x + 6y = 9 (for x) Ex. F = 9C + (for C) 5 (Rethink how to write the equation.) y 1x + ( - 6y) = 18 x x = 9 6y C F = 9C 5-6y = 18 1x * x = 9-6y *9 * - 6 5(F ) = 9C y = 18-1x + 6y / 5-6 5(F ) = C + 1x + 9 y = - + x y = x (Convert Celsius to Fahrenheit)

(Equation of an ellipse) (Equation for the surface area of a sphere.) (Equation for the volume of a sphere) Ex. x + 7y = 6 (for y) Ex. 5 A = π r (for r) Ex. 6 V = π r (for r) 7 y 7y = 6 x r A = r r V = π r π y y = 6 x r r V = r A 7 r = π * 7 p *π *π 6 - x y = ± V 7 r = + x / p ( Since y could be positive or negative, it is necessary to include + symbol with the square root operation. ) ( Since r is always positive, it is not necessary to include + symbol with the square root operation. ) ( Never include a + symbol when reversing the cube of a number operation. ) Ex. 7 P + Prt = 5,000 (for P) Ex. 8 a = a + 1 (for a) Ex. 9 - = 1 (for b) b c a b c (Get P by itself by factoring out P.) P(1 + rt) = 5,000 ac c = ab + b P = 5,000 1 + rt Comment: Make sure that you completely understand the difference between examples 7 and 10. (Gather a on the same side of the equation.) ac ab c = b (Move -c to the other side of the equation..) ac ab = c + b a(c b) = c + b a = c + b c b (The variable a is now gathered on the left side of the equation.. ) (Get a by itself by factoring out a.) 1 abc æ ç - ö = abc æ ç ö è a b ø è c ø bc - ac = ab bc - ab = ac b( c - a) = ac ac b = c - a (Remove fractions by multiplying both sides by the lowest common denominator.) (Gather the variable b on the left side of the equation.. ) (Factor out b. ) Ex. 10. P + Prt = 5,000 (for r) Ex. 11. y - y = 10 (for y) 5 x + r Prt = 5,000 P æ y y ö 5( x + ) ç - = 5( x + )10 * Pt r = 5,000 - P è 5 x + ø Pt + P y( x + ) - 5y = 50( x + ) xy + 6y - 5y = 50x + 100 y(x + 1) = 50x + 100 50x + 100 y = x + 1 xy + y = 50x + 100 (The variable y is now gathered on the left side of the equation.. )

Ex. 1. Consider the implicitly defined relation x xy = y -. The graph of this relation is shown below. We will rewrite the relation as a function of x and then as two functions of y. The graphs of all of the equations are identical! 8 Rewrite as a function of x by solving for y. Rewrite as a function of y by solving for x. x + = xy + y x + = y(x + ) y = x + x + Gather y on the right side of equation. Factor out common factor of y. Since there is a x and x in the equation, we will use the quadratic formula to solve for x. x xy -y + = 0 a =, b = -y, and c = -y + Set equation equal to zero. Identify a, b, and c. Observations: The above formula expresses y as a function of x. The domain of the function equals all real numbers except x = -/. The estimated range is (y < -.0) or (y > 0.). The numerator is always positive. Therefore the function has no real roots and the graph of the function can not intersect the x-axis. When x = -/, the denominator equals zero. Therefore the line x = -/ is a vertical asymptote of the graph.. As x gets closer to -/ the function values approach plus or minus infinity. When x = -/ +.0001, y = 18,516.7. When x = -/ -.0001, y = -18,50.0 x = x = y ± (- y) - ( - 8y + ) y ± 9y + y -16 Observations: Plug a, b, and c into the quadratic formula. Simplify expression. Because of the + operator, the above formula expresses x in terms of y, but not as a single function of y. An input value for y will generally have two outputs. Recall that if a relation is also a function, then every in input value to the function can have no more than one output value. The estimated domain of this relation is (y < -.0) or (y > 0.). How can anyone not be impressed by the power of the quadratic formula!

These equations can be easily solved by applying simple properties of fractions, factoring and old fashioned thinking! 9 1. x + = x -1 x -1 5. z 1 + = z + 1 z + 1. 5 1 - = 8 + a a a 6. A mathematical proof that 1 =. Can you find the mistake? Let a and b be any two real numbers that are equal. a = b (So we let a equal b.) a = ab (Multiply both sides by a ) a - b = ab - b (Subtract b from both sides.). x 5 - = x - 5 x - 5 (a + b)(a b) = b(a b) ( a + b)( a - b) b( a - b) = ( a - b) ( a - b) (Factor both sides) (Divide both sides by a-b ) a + b = b (Simplify both sides.) b + b = b (Substitute b for a) b = b (Simplify left side) b b = b b (Divide both sides by b. ). a - = a a + a + = 1 (Simplify both sides of the quation.)