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Solve each of the following equations. Round the result to the nearest tenth if necessary. a. \(3 x=15.3\) b. \(-2.3 x=10.35\) c. \(-5.2 a=-44.2\) d. \(15.2=15.2 y\) e. \(98.8=-4 y\) f. \(4.2 x=-\sqrt{64}\) g. \(-0.6 x+2.1 x=54\) h. \(4.675 x-(-3.334 x)=-18.50079\) i. \(2.8 x-3.08 x=-0.294\)

Short Answer

Expert verified
Question: Solve the following linear equations: a) \(3x = 15.3\) b) \(-2.3x = 10.35\) c) \(-5.2a = -44.2\) d) \(15.2 = 15.2y\) e) \(98.8 = -4y\) f) \(4.2x = -\sqrt{64}\) g) \(-0.6x + 2.1x = 54\) h) \(4.675x - (-3.334x) = -18.50079\) i) \(2.8x - 3.08x = -0.294\)

Step by step solution

01

Problem a: Solve \(3x=15.3\)

To solve this equation, divide both sides by 3: \(x = \frac{15.3}{3}\) \(x \approx 5.1\)
02

Problem b: Solve \(-2.3x=10.35\)

To solve this equation, divide both sides by \(-2.3\): \(x = \frac{10.35}{-2.3}\) \(x \approx -4.5\)
03

Problem c: Solve \(-5.2a=-44.2\)

To solve this equation, divide both sides by \(-5.2\): \(a = \frac{-44.2}{-5.2}\) \(a \approx 8.5\)
04

Problem d: Solve \(15.2=15.2y\)

To solve this equation, divide both sides by \(15.2\): \(y = \frac{15.2}{15.2}\) \(y = 1\)
05

Problem e: Solve \(98.8=-4y\)

To solve this equation, divide both sides by \(-4\): \(y = \frac{98.8}{-4}\) \(y \approx -24.7\)
06

Problem f: Solve \(4.2x=-\sqrt{64}\)

To solve this equation, first find the square root of 64: \(\sqrt{64} = 8\) Now, divide both sides by \(4.2\): \(x = \frac{-8}{4.2}\) \(x \approx -1.9\)
07

Problem g: Solve \(-0.6x+2.1x=54\)

To solve this equation, first combine the like terms: \(1.5x=54\) Next, divide both sides by \(1.5\): \(x = \frac{54}{1.5}\) \(x = 36\)
08

Problem h: Solve \(4.675x-(-3.334x)=-18.50079\)

To solve this equation, first combine the like terms: \(8.009x=-18.50079\) Next, divide both sides by \(8.009\): \(x = \frac{-18.50079}{8.009}\) \(x \approx -2.311\)
09

Problem i: Solve \(2.8x-3.08x=-0.294\)

To solve this equation, first combine the like terms: \(-0.28x=-0.294\) Next, divide both sides by \(-0.28\): \(x = \frac{-0.294}{-0.28}\) \(x \approx 1.05\)

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