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Prepared by Doron Shahar Linear Equations Prepared by Doron Shahar Warm-up A linear equation in one variable is an equation that can be written in ax b 0 where a and b are constants and a ≠ 0. the form __________ Solve for x in the following equations. 3x 6 0 2 x 5 2 x 9 x 3 1 x x 12 2 5 x 3 ( x 1) Prepared by Doron Shahar A simple linear equation Starting Equation 3x 6 Divide both sides of the equation by 3 3x Solution 3 6 3 x2 Prepared by Doron Shahar General method for solving equations of the form ax c Starting Equation ax c Divide both sides of the equation by a ax Solution c a a x c a eg. eg. 3x 6 3x 3 eg. 6 3 x2 Prepared by Doron Shahar Example 1 Starting Equation 2 x 9 Divide both sides of the equation by 2 2x Solution 2 9 2 x 9 2 Prepared by Doron Shahar Example 2 Starting Equation x 12 Divide both sides of the equation by −1 x Solution 1 12 1 x 12 Prepared by Doron Shahar Grouping like terms Starting Equation 0 2 x 5 2x 2x 2x 5 Divide both sides of the equation by 2 Solution 2x 5 2 2 x 5 2 Prepared by Doron Shahar Grouping like terms The idea behind “grouping like terms” is to get all of the terms involving x on one side of the equation and all of the constants on the other side of the equation. Take the previous problem as an example. Starting Equation 0 2 x 5 2x Grouping like terms Term involving 2x 2x 5 x Constant Prepared by Doron Shahar Example: Grouping like terms Starting Equation x 3 1 x 3 Group like terms 3 x 4 x x x 2x 4 Divide both sides of the equation by −2 Solution 2x 2 x 2 4 2 Prepared by Doron Shahar Solving Linear Equations Using Distribution Starting Equation 2 5 x 3 ( x 1) Distribute 2 5x 3x 3 2 Group like terms Divide both sides of the equation by −8 Solution 2 5x 3x 1 3x 3x 8x 1 8 8 x 1 8 Prepared by Doron Shahar 1.1.2(A) Starting Equation 4 ( x 1) 2 3 ( 2 x 1) 4 Distribute 4x 4 2 6x 3 4 4x 2 6x 7 2 Groups like terms Divide both sides by −2 Solution 2 4x 6x 5 6x 6x 2 x 5 2 2 x 5 2 Prepared by Doron Shahar Overview Our method of solving linear equations involves repeated simplification. Each step is designed to reduce the problem to a form that we already now how to solve. Steps: 1. Distributing: Results in only needing to group like terms 2. Grouping like terms: Results in the form 3. Dividing: Results in a solution ax c Prepared by Doron Shahar Reducing to a previously solved problem TELL JOKE NOW!! Next, we will learn how to solve linear equations with • Decimals • Fractions • Variables in the denominator. Rather than solving such problems, I will teach you how to reduce them to previously solved problems. That is, I will show you how to convert such problems into the form we just learned how to solve. Prepared by Doron Shahar Linear equations with Decimals If a linear equation has decimals, we multiply both sides of the equation by a power of 10 (i.e., 10, 100, 1000, etc) to get rid of the decimals. This is not necessary, but it can help if you don’t like working with decimals. Example: If our problem has the decimals 0.1, 0.2, and −0.8, we multiply both sides of the equation by 10. Our decimals then become 10×0.1=1, 10 ×0.2=2, and 10 ×(−0.8)= −8 Prepared by Doron Shahar 1.1.2(C) Linear equation with decimals Starting Equation Multiply by 10 to get rid of decimals Distribute the 10 0 . 1( 4 x 8 ) 0 . 2 ( x 4 ) 0 . 8 x 10 0 . 1( 4 x 8 ) 0 . 2 ( x 4 ) 10 0 . 8 x 10 0 . 1( 4 x 8 ) 10 0 . 2 ( x 4 ) 10 0 . 8 x 1( 4 x 8 ) 2 ( x 4 ) 8 x The problem is now in a form you can solve. Prepared by Doron Shahar Example: Linear equation with decimals Starting Equation Multiply by 100 to get rid of decimals Distribute the 100 0 . 03 ( x 1) 0 . 17 x 0 . 01 x 100 0 . 03 ( x 1) 0 . 17 x 100 0 . 01 x 100 0 . 03 ( x 1) 100 0 . 17 x 100 0 . 01 x 3 ( x 1) 17 x x The problem is now in a form you can solve. Prepared by Doron Shahar Linear equations with Fractions If a linear equation has fractions, we multiply both sides of the equation by a common denominator to get rid of the fractions. Ideally, we should multiply by the least common denominator. This is not necessary, but it can help if you don’t like working with fractions. Example: If our problem has the fractions 1/4, 1/2, and 1/8, we multiply both sides of the equation by 8. Our fractions then become 8×1/4=2, 8 ×1/2=4, and 8 ×1/8= 1. Prepared by Doron Shahar 1.1.2(B) Linear equation with fractions Starting Equation Multiply by 8 to get rid of fractions Distribute the 8 1 4 ( p 2) 1 2 ( 3 p 1) 1 ( p 1) 8 1 1 1 8 ( p 2 ) ( 3 p 1) 8 ( p 1) 2 4 8 1 1 1 8 ( p 2 ) 8 ( 3 p 1) 8 ( p 1) 4 2 8 2 ( p 2 ) 4 ( 3 p 1) ( p 1) The problem is now in a form you can solve. Prepared by Doron Shahar Example: Linear equation with fractions Starting Equation Multiply by 15 to get rid of fractions Distribute the 15 1 3 ( 2 y 1) 1 15 ( y 4) 1 ( y 1) 5 1 1 1 15 ( 2 y 1) ( y 4 ) 15 ( y 1) 15 3 5 1 1 1 15 ( 2 y 1) 15 ( y 4 ) 15 ( y 1) 3 15 5 5 ( 2 y 1) 1( y 4 ) 3 ( y 1) The problem is now in a form you can solve. Prepared by Doron Shahar Variables in the denominators If an equation has variables in the denominators, it is NOT a linear equation. Such equations, however, can lead to linear equations. We treat such equations like those with fractions. That is, we multiply both sides of the equation by a common denominator to get rid of the variables in the dominators. Ideally, we should multiply by the least common denominator. Example: If our problem has x−3 in the denominator of one term, and x−3 in the denominator of another term, we multiply both sides of the equation by x−3. After the multiplication, the terms will have no variables in the denominators. Prepared by Doron Shahar 1.1.2(D) Variables in the denominators Starting Equation Multiply both sides by (x −3) Distribute the (x−3) 2x x3 1 5 x3 2x 5 ( x 3) 1 ( x 3 ) x3 x 3 2x 5 ( x 3) ( x 3) 1 ( x 3) x 3 x 3 2 x ( x 3) 5 The problem is now in a form you can solve. Prepared by Doron Shahar Example: With variables in in the denominators Starting Equation 3 x 1 2 x2 4 ( x 1)( x 2 ) Multiply both sides by (x −1)(x+2) 2 4 3 ( x 1)( x 2 ) ( x 1)( x 2 ) x 1 x 2 ( x 1 )( x 2 ) Distribute the (x−1)(x+2) 4 3 2 ( x 1)( x 2 ) ( x 1)( x 2 ) ( x 1)( x 2 ) x 1 x 2 ( x 1 )( x 2 ) 3 ( x 2 ) 2 ( x 1) 4 The problem is now in a form you can solve. Prepared by Doron Shahar Example: With variables in in the denominators Starting Equation 3 x 1 2 x2 4 x x2 2 The denominator x2+x−2 on the right-hand-side of the equation is not factored. That makes it difficult to find the least common denominator. Therefore, we first factor x2+x−2. Equation after factoring 3 x 1 2 x2 4 ( x 1)( x 2 ) That is the starting equation on the previous slide. Prepared by Doron Shahar Extraneous Solutions Starting Equation x x 1 Multiply both sides by (x−1) Distribute the (x−1) 1 1 x 1 x 1 ( x 1) 1 ( x 1) x 1 x 1 x 1 ( x 1) ( x 1) 1 ( x 1) x 1 x 1 x ( x 1) 1 Extraneous Solution x 1 Plugging in 1 for x in the original equation would result in dividing by zero. Thus, x=1 is not a solution. It is called an extraneous solution. Prepared by Doron Shahar Variables in the denominators There are two key differences between equations with variables in the denominators and those with fractions. 1. If an equation has variables in the denominators, it is important to first factor the expressions in the denominators. This makes it easier to find a least common denominator. 2. An equation with variables in the denominator may have extraneous solutions. Therefore, it is important to check that the solution you get does not make any of the denominators zero. Checking answers on a Calculator 1. Type in your answer. (e.g. 4) 2. Press STO, X,T,Ѳ,n , and press ENTER 3. Enter the original equation. (e.g. 2x+1=9) (Use X,T,Ѳ,n to type x. Press 2nd, then MATH (TEST), then ENTER to type =.) 4. Press ENTER. If a 1 appears your answer is right. If a 0 appears your answer is wrong. Prepared by Doron Shahar Identifying Linear equations A linear equation in one variable is an equation that can be written in ax b 0 where a and b are constants and a ≠ 0. the form __________ To decide if an equation is linear, we try and get it in this form. To do this, simply try to solve it. But instead of actually solving it, stop short to try and get it in this form. If it cannot be written in this form, then the equation is not linear. Note: When trying to get the equation into this form, you cannot multiply or divide by an expression with a variable. Prepared by Doron Shahar Example: Identifying linear equations Starting Equation 7 x 1 3x 3x 3x Equation in the form ax b 0 4x 1 0 b 1 ax b 0 a4 The equation is linear. Prepared by Doron Shahar 1.1.1(B) Identifying linear equations Starting Equation 4x 3 x 2x 3 x x Equation in the form ax b 0 a 1 b 3 x 1x 3 0 ax b 0 The equation is linear. Prepared by Doron Shahar 1.1.1(D) Identifying linear equations Starting Equation ( x 1) x 7 Note: When trying to get the equation into the form ax+b=0, you cannot multiply or divide by an expression with a variable. We would have to multiply by x to get the equation into the from ax+b=0. Therefore, the equation is non-linear. It does, however, lead to the linear equation −6x−1=0. The equation is non-linear. Prepared by Doron Shahar 1.1.1(C) Identifying linear equations Starting Equation 3x 1 x 2 Note: When trying to get the equation into the form ax+b=0, you cannot multiply or divide by an expression with a variable. We would have to multiply by x to get the equation into the from ax+b=0. Therefore, the equation is non-linear. It also does NOT lead to a linear equation. The equation is non-linear. Prepared by Doron Shahar 1.1.1(A) Identifying linear equations Starting Equation 3x 2 4x 2 4x 2 4x 2 4x 3x 2 0 2 The equation cannot be written in the form ax The equation is non-linear. b 0 Prepared by Doron Shahar Quadratic equations An equation in the form ax2+bx+c=0 where a≠0 is called a quadratic equation. The equation −4x2+3x+2=0 from the previous slide is an example of a quadratic equation. All quadratic equations are non-linear. You may simply assume this fact. Although a mathematician would want you to prove it. JOKE: An argument is needed to convince a reasonable person. A proof is needed to convince and unreasonable one. Prepared by Doron Shahar Systems of equations So far we have learned to solve for one variable given a single equation. Next we shall try and solve for multiple variables given multiple equations. Examples: 2 x 5 y 7 x 2y 3 x 2y 14 x 14 y 1 There are two methods for solving systems of equations: Substitution and Elimination. Both work by combining the equations into a single equation with one variable. Prepared by Doron Shahar Substitution Starting Equations x 2y 14 x 14 y 1 Substitute 2y for x in the second equation 14 ( 2 y ) 14 y 1 Solution for y Plug in 1/42 for y in the first equation to get the solution for x y 1 42 1 1 x 2 21 42 Prepared by Doron Shahar 1.1.3 Substitution Starting Equations Solve for x in the second equation Substitute −2y−8 for x in the first equation Solution for y Plug in −5 for y in the second equation to get an equation for x 4 x 2 y 18 x 2 y 8 x 2 y 8 4 ( 2 y 8 ) 2 y 18 y 5 x 2( 5) 8 Prepared by Doron Shahar 1.1.3 Elimination 4x 2 y Starting Equations x 2y Add the two equations together Solution for x Plug in 2 for x in one of the original equations to get an equation for y 18 8 5 x 10 x 2 2 2 y 8 Now solve for y Prepared by Doron Shahar 1.1.4 Elimination Starting Equations Multiply the first equation by −2 Add the latter two equations together 3 A 5 B 15 6 A 2 B 12 6 A 10 B 30 12 B 18 We chose to multiply the first equation by −2, so that when we would add the two equations together the terms with A would cancel leaving us with an equation that has just one variable. Prepared by Doron Shahar Substitution and elimination Both substitution of elimination work by combining the equations into a single equation with one variable. Substitution accomplishes this by substituting one variable by an equivalent expression that is written in terms of another variable. Elimination works by multiplying the equations by appropriate constants, so that when added together one of the variables will be eliminated. The resulting equation with one variable is then solved. The solution to this equation is plugged into one of the original equations . That produces an equation in the second variable, which may then be solved.