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43literal equations

The document discusses solving literal equations by isolating the variable of interest on one side of the equation. It provides examples of solving equations for various variables by adding, subtracting, multiplying, or dividing both sides of the equation by the same quantity. The goal is to isolate the variable being solved for so it stands alone on one side of the equal sign. Steps include clearing fractions, moving all other terms to the other side of the equation, and then dividing both sides by the coefficient of the variable being solved for.

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Literal Equations
Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. Literal Equations
Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. Literal Equations Example A. a. Solve for x if x + b = c b. Solve for w if yw = 5.
Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations. Example A. a. Solve for x if x + b = c Literal Equations b. Solve for w if yw = 5.
Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations. Example A. a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b Literal Equations b. Solve for w if yw = 5.
Example A. a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b Literal Equations b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.
Example A. a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b Literal Equations b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. yw = 5 yw/y = 5/y Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.
Example A. a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b 5 y Literal Equations b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. yw = 5 yw/y = 5/y w = Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.
Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Example B. a. Solve for x if (a + b)x = c Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Example B. a. Solve for x if (a + b)x = c (a + b) x = c div the RHS by (a + b) Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
To solve for a specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Example B. a. Solve for x if (a + b)x = c (a + b) x = c x = c (a + b) div the RHS by (a + b) Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
b. Solve for w if 3y2w = t – 3 Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 div the RHS by 3y2 Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 w = t – 3 3y2 div the RHS by 3y2 Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 w = t – 3 3y2 div the RHS by 3y2 Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 w = t – 3 3y2 div the RHS by 3y2 move b2 to the RHS Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 div the RHS by 3y2 move b2 to the RHS Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax – ay = 10 – ax
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax – ay = 10 – ax div by –a
b. Solve for w if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax – ay = 10 – ax div by –a y = 10 – ax –a
Multiply by the LCD to get rid of the denominator then solve. Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d– 4 = d 3d + b Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b move –4d and b Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b – b = 3d + 4d move –4d and b Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b – b = 3d + 4d move –4d and b – b = 7d Example C. Solve for d if Literal Equations
–4 = d 3d + b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b – b = 3d + 4d move –4d and b – b = 7d = d 7 –b Example C. Solve for d if Literal Equations div. by 7
Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. Literal Equations x = 3 + 1 2y Example D. Solve for y if
Flipping Equations Literal Equations
Flipping Equations Literal Equations x = 3 + 1 2y Example D. Solve for y if
Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. Literal Equations x = 3 + 1 2y Example D. Solve for y if
= D C Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: Literal Equations B A = C D A B Reciprocate both sides x = 3 + 1 2y Example D. Solve for y if
= D C Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y Reciprocate both sides
= D C Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y Reciprocate both sides
= D C Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y reciprocating both sides, = 1 2y x – 3 1 Reciprocate both sides
= D C Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y reciprocating both sides, then div. by 2 = 1 2y x – 3 1 = 2y 2(x – 3) 1 Reciprocate both sides 2
= D C Flipping Equations If it’s advantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y reciprocating both sides, then div. by 2 = 1 2y x – 3 1 = 2y 2(x – 3) 1 Reciprocate both sides 2 = y or 2(x – 3) 1 y =
Exercise. Solve for the indicated variables. Literal Equations 1. a – b = d – e for b. 2. a – b = d – e for e. 3. 2*b + d = e for b. 4. a*b + d = e for b. 5. (2 + a)*b + d = e for b. 6. 2L + 2W = P for W 7. (3x + 6)y = 5 for y 8. 3x + 6y = 5 for y w = t – 3 613. for t w = t – b a14. for t w =11. for t w =12. for t t 6 6 t w = 3t – b a15. for t 16. (3x + 6)y = 5 for x w = t – 3 6 17. + a for t w = at – b 5 18. for t w = at – b c 19. + d for t 3 = 4t – b t 20. for t 9. 3x + 6xy = 5 for y 10. 3x – (x + 6)y = 5z for y

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byRAKESH SAJJAN
 

43literal equations

  • 1.
  • 2.
    Given an equationwith many variables, to solve for a particular variable means to isolate that variable to one side of the equation. Literal Equations
  • 3.
    Given an equationwith many variables, to solve for a particular variable means to isolate that variable to one side of the equation. Literal Equations Example A. a. Solve for x if x + b = c b. Solve for w if yw = 5.
  • 4.
    Given an equationwith many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations. Example A. a. Solve for x if x + b = c Literal Equations b. Solve for w if yw = 5.
  • 5.
    Given an equationwith many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations. Example A. a. Solve for x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b Literal Equations b. Solve for w if yw = 5.
  • 6.
    Example A. a. Solvefor x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b Literal Equations b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.
  • 7.
    Example A. a. Solvefor x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b Literal Equations b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. yw = 5 yw/y = 5/y Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.
  • 8.
    Example A. a. Solvefor x if x + b = c Remove b from the LHS by subtracting from both sides x + b = c –b –b x = c – b 5 y Literal Equations b. Solve for w if yw = 5. Remove y from the LHS by dividing both sides by y. yw = 5 yw/y = 5/y w = Given an equation with many variables, to solve for a particular variable means to isolate that variable to one side of the equation. We try to accomplish this just as solving equations for the x, by +, – the same term, or * , / by the same quantity to both sides of the equations.
  • 9.
    Literal Equations Adding orsubtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 10.
    To solve fora specific variable in a simple literal equation, do the following steps. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 11.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 12.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 13.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 14.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 15.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Example B. a. Solve for x if (a + b)x = c Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 16.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Example B. a. Solve for x if (a + b)x = c (a + b) x = c div the RHS by (a + b) Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 17.
    To solve fora specific variable in a simple literal equation, do the following steps. 1. If there are fractions in the equations, multiple by the LCD to clear the fractions. 2. Isolate the term containing the variable we wanted to solve for – move all the other terms to other side of the equation. 3. Isolate the specific variable by dividing the rest of the factor to the other side. Literal Equations Example B. a. Solve for x if (a + b)x = c (a + b) x = c x = c (a + b) div the RHS by (a + b) Adding or subtracting a term to both sides may be viewed as moving the term across the " = " and change its sign.
  • 18.
    b. Solve forw if 3y2w = t – 3 Literal Equations
  • 19.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 div the RHS by 3y2 Literal Equations
  • 20.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 w = t – 3 3y2 div the RHS by 3y2 Literal Equations
  • 21.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 w = t – 3 3y2 div the RHS by 3y2 Literal Equations
  • 22.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 w = t – 3 3y2 div the RHS by 3y2 move b2 to the RHS Literal Equations
  • 23.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 div the RHS by 3y2 move b2 to the RHS Literal Equations
  • 24.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations
  • 25.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations
  • 26.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10
  • 27.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand
  • 28.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10
  • 29.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax
  • 30.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax – ay = 10 – ax
  • 31.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax – ay = 10 – ax div by –a
  • 32.
    b. Solve forw if 3y2w = t – 3 3y2w = t – 3 c. Solve for a if b2 – 4ac = 5 b2 – 4ac = 5 – 4ca = 5 – b2 w = t – 3 3y2 a = 5 – b2 –4c div the RHS by 3y2 move b2 to the RHS div the RHS by –4c Literal Equations d. Solve for y if a(x – y) = 10 a(x – y) = 10 expand ax – ay = 10 subtract ax – ay = 10 – ax div by –a y = 10 – ax –a
  • 33.
    Multiply by theLCD to get rid of the denominator then solve. Literal Equations
  • 34.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. Example C. Solve for d if Literal Equations
  • 35.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d– 4 = d 3d + b Example C. Solve for d if Literal Equations
  • 36.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d Example C. Solve for d if Literal Equations
  • 37.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d Example C. Solve for d if Literal Equations
  • 38.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b Example C. Solve for d if Literal Equations
  • 39.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b move –4d and b Example C. Solve for d if Literal Equations
  • 40.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b – b = 3d + 4d move –4d and b Example C. Solve for d if Literal Equations
  • 41.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b – b = 3d + 4d move –4d and b – b = 7d Example C. Solve for d if Literal Equations
  • 42.
    –4 = d 3d +b Multiply by the LCD to get rid of the denominator then solve. multiply by the LCD d d – 4 = d 3d + b – 4 = d 3d + b ( ) d – 4d = 3d + b – b = 3d + 4d move –4d and b – b = 7d = d 7 –b Example C. Solve for d if Literal Equations div. by 7
  • 43.
    Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. Literal Equations x = 3 + 1 2y Example D. Solve for y if
  • 44.
  • 45.
    Flipping Equations Literal Equations x= 3 + 1 2y Example D. Solve for y if
  • 46.
    Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. Literal Equations x = 3 + 1 2y Example D. Solve for y if
  • 47.
    = D C Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: Literal Equations B A = C D A B Reciprocate both sides x = 3 + 1 2y Example D. Solve for y if
  • 48.
    = D C Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y Reciprocate both sides
  • 49.
    = D C Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y Reciprocate both sides
  • 50.
    = D C Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y reciprocating both sides, = 1 2y x – 3 1 Reciprocate both sides
  • 51.
    = D C Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y reciprocating both sides, then div. by 2 = 1 2y x – 3 1 = 2y 2(x – 3) 1 Reciprocate both sides 2
  • 52.
    = D C Flipping Equations If it’sadvantageous to do so, we may reposition the equation by reciprocating both sides of an equation. In particular: x = 3 + 1 2y Example D. Solve for y if Literal Equations B A = C D A B x = 3 + 1 2y x – 3 = 1 2y reciprocating both sides, then div. by 2 = 1 2y x – 3 1 = 2y 2(x – 3) 1 Reciprocate both sides 2 = y or 2(x – 3) 1 y =
  • 53.
    Exercise. Solve forthe indicated variables. Literal Equations 1. a – b = d – e for b. 2. a – b = d – e for e. 3. 2*b + d = e for b. 4. a*b + d = e for b. 5. (2 + a)*b + d = e for b. 6. 2L + 2W = P for W 7. (3x + 6)y = 5 for y 8. 3x + 6y = 5 for y w = t – 3 613. for t w = t – b a14. for t w =11. for t w =12. for t t 6 6 t w = 3t – b a15. for t 16. (3x + 6)y = 5 for x w = t – 3 6 17. + a for t w = at – b 5 18. for t w = at – b c 19. + d for t 3 = 4t – b t 20. for t 9. 3x + 6xy = 5 for y 10. 3x – (x + 6)y = 5z for y