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If you can t see straightaway how these solutions are derived, Tables 12-6 through 12-10 show how the equations can be solved, step-by-step Note that in the third and fifth original equations above (and in Tables 12-8 and 12-10), we must not let a equal 0 Also, in the fourth solution equation (and in Table 12-9), we must never allow either a or b to equal 0
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Table 12-6
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Statements 4x = 0 4x /4 = 0/4 x=0
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Process for solving the equation 4x = 0
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Reasons This is the equation we are given Divide each side through by 4 Simplify each side
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Table 12-7
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Statements x /7 = 2 7x /7 = 7 2 x = 14
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Reasons This is the equation we are given Multiply each side by 7 Simplify each side
Table 12-8
Statements 2x /a = b a(2x /a) = ab 2x = ab 2x /2 = ab /2 x = ab /2
Process for solving the equation 2x /a = b, provided a 0
Reasons This is the equation we are given Multiply each side by a Simplify the left side Divide each side by 2 Simplify the left side
Table 12-9
Process for solving the equation 5abx = c, provided a 0 and b 0
Reasons This is the equation we are given We are about to divide through by the product of these constants This will allow us to solve the equation in a streamlined fashion Divide through by the constant (5ab) Simplify the left side
Statements 5abx = c Require that a 0 and b 0 Consider (5ab) to be a single constant 5abx /(5ab) = c /(5ab) x = c /(5ab)
Table 12-10
Process for solving the equation 3x /(4a) = 3, provided a 0
Reasons This is the equation we are given This will allow us to solve the equation in a streamlined fashion Multiply through by the constant (4a) Simplify both sides Divide each side by 3 Simplify each side
Statements 3x /(4a) = 3 Consider (4a) to be a single constant [3x /(4a)](4a) = 3 (4a) 3x = 12a (3x)/3 = 12a /3 x = 4a
198 First-Degree Equations in One Variable
Here s a challenge!
Manipulate the following equation so it contains x all by itself on the left side, and an expression containing the constants without x on the right side Indicate, if applicable, which constants cannot equal 0 3abx /(4cd ) = k 2
Solution
We must have c 0 and d 0 because, if either of them are allowed to equal 0, the left-hand side of the equation becomes undefined Let s multiply the equation through by the quantity (4cd) We get [3abx /(4cd )](4cd ) = (4cd )k 2 which simplifies to 3abx = 4cdk 2 Now we can divide the entire equation through by the quantity (3ab) When we do this, we must insist that a 0 and b 0 That produces 3abx /(3ab) = 4cdk 2/(3ab) which simplifies to x = 4cdk 2/(3ab) That does it! We don t have to worry about the fact that one of the constants is squared The square of a constant is always another constant The variable, x, is never raised to any power (other than the first power), so the equation is a first-degree equation
Combinations of Operations
In a first-degree equation that involves a single variable, constants can be added to or subtracted from that variable, and the variable can also be multiplied or divided by nonzero constants
Examples Here are some first-degree equations that involve combinations of sums, differences, products, and ratios:
8x 4 = 0 18x + 7 = 2 a 3x = 0 a 5 + 15x = 0
Combinations of Operations
a 8x = b 6a + 3x = 12b 6a 3x /(bc) = 24d These seven equations can all be rearranged with the morphing laws we already know, so that x appears alone on the left sides of the equality symbols, and nothing but constants appear on the right sides Here are the respective solutions: x = 1/2 x = 1/2 x = a /3 x = 1/3 a /15 x = a /8 b /8 x = 2a + 4b x = 2abc + 8bcd
Are you confused
Tables 12-11 through 12-17 break down the solution processes for the above equations Some of the steps are combined, making the derivations less tedious than those earlier in this chapter Note that in the last original equation above (and in Table 12-17), it s necessary that b 0 and c 0 Also note that an attempt has been made to put the solutions in elegant form by avoiding sums or differences in the numerators of fractions, putting fractions in lowest terms, and getting the letters for the constants in alphabetical order
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