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3 years ago

What is the solution set to the equation (3x−9)(5x−3)=0?

2 answers:

8 0

This product could be zero if one or both of the parenthesis are zero.
3x-9=0; 3x=9; x= 3
5x-3=0; 5x=3; x=3/5

The solution is { 3/5, 3}

almond37 [142]3 years ago

6 0

Answer:

({3,\frac{3}{5} })

Step-by-step explanation:

Given is an equation in x with two factors on left side and 0 on right side as

$(3x−9)(5x−3)=0$

We know that when product of two factors is zero, then any one of the factors can be 0

Equate each factor to 0

$(3x−9)=0\\(5x−3)=0\\x=3\\x=\frac{3}{5}$

Thus we have two solutions

x=3\\

x=\frac{3}{5}

Solution set is

$({3,\frac{3}{5} })$

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mixer [17]

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3 years ago

Solve. clear fraction or decimals first . 2/5x-1/3x=5

mafiozo [28]

2/5x-1/3x=5

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3 0

4 years ago

Solve using the Quadratic Formula 5x2-3x-7=0 A) B) C) D)

vlabodo [156]

Answer:

C) $\displaystyle x=\frac{3 \pm \sqrt{149}}{10}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Standard Form: ax² + bx + c = 0
Quadratic Formula: $\displaystyle x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

Step 1: Define

Identify

5x² - 3x - 7 = 0

↓

a = 5, b = -3, c = -7

Step 2: Solve for x

Substitute in variables [Quadratic Formula]: $\displaystyle x=\frac{3 \pm \sqrt{(-3)^2-4(5)(-7)}}{2(5)}$
[√Radical] Evaluate exponents: $\displaystyle x=\frac{3 \pm \sqrt{9-4(5)(-7)}}{2(5)}$
[√Radical] Multiply: $\displaystyle x=\frac{3 \pm \sqrt{9+140}}{2(5)}$
[√Radical] Add: $\displaystyle x=\frac{3 \pm \sqrt{149}}{2(5)}$
[Fraction] Multiply: $\displaystyle x=\frac{3 \pm \sqrt{149}}{10}$