Math / Algebra

QuestionSolve the proportion. Express numbers as integers or simplified fractions. $\frac{1}{x-3}=\frac{x-3}{4}$
The solution set is $\square$ \}.

Studdy Solution

STEP 1

What is this asking? We need to find the values of $x$ that make the given proportion true. Watch out! Remember that we can't divide by zero, so we need to make sure that any solutions we find don't make the denominator zero.
Also, remember that when we take the square root of both sides of an equation, we need to consider both the positive and negative square roots.

STEP 2

1. Cross-multiply
2. Expand and rearrange
3. Solve the quadratic equation

STEP 3

Alright, let's cross-multiply!
We multiply the numerator of the left side by the denominator of the right side, and the denominator of the left side by the numerator of the right side.
This is valid because if two fractions are equal, their cross-products are equal.
This gives us: $1 \cdot 4 = (x-3) \cdot (x-3)$

STEP 4

This simplifies to: $4 = (x-3)^2$

STEP 5

Now, let's expand the right side of the equation: $4 = x^2 - 6x + 9$

STEP 6

To solve this quadratic equation, we want to set it equal to zero.
Let's subtract 4 from both sides: $0 = x^2 - 6x + 5$

STEP 7

We can factor this quadratic equation!
We are looking for two numbers that multiply to 5 and add up to -6.
Those numbers are -5 and -1.
So, we can rewrite the equation as: $(x-5)(x-1) = 0$

STEP 8

Now, we use the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.
This gives us two possible solutions: $x-5 = 0 \quad \text{or} \quad x-1 = 0$

STEP 9

Solving these two equations gives us $x = 5$ and $x = 1$ .
Let's check if these solutions make the original denominators zero.
If we substitute $x = 5$ into the original equation, we get $\frac{1}{5-3} = \frac{5-3}{4}$ , which simplifies to $\frac{1}{2} = \frac{2}{4}$ , or $\frac{1}{2} = \frac{1}{2}$ .
That works!
If we substitute $x = 1$ into the original equation, we get $\frac{1}{1-3} = \frac{1-3}{4}$ , which simplifies to $\frac{1}{-2} = \frac{-2}{4}$ , or $-\frac{1}{2} = -\frac{1}{2}$ .
That works too!

STEP 10

The solution set is $\{1, 5\}$ .

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QuestionSolve the proportion. Express numbers as integers or simplified fractions. $\frac{1}{x-3}=\frac{x-3}{4}$
The solution set is $\square$ \}.

Studdy Solution

STEP 1

STEP 2

1. Cross-multiply
2. Expand and rearrange
3. Solve the quadratic equation

STEP 3

STEP 4

This simplifies to: $4 = (x-3)^2$

STEP 5

Now, let's expand the right side of the equation: $4 = x^2 - 6x + 9$

STEP 6

To solve this quadratic equation, we want to set it equal to zero.
Let's subtract 4 from both sides: $0 = x^2 - 6x + 5$

STEP 7

We can factor this quadratic equation!
We are looking for two numbers that multiply to 5 and add up to -6.
Those numbers are -5 and -1.
So, we can rewrite the equation as: $(x-5)(x-1) = 0$

STEP 8

Now, we use the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.
This gives us two possible solutions: $x-5 = 0 \quad \text{or} \quad x-1 = 0$

STEP 9

STEP 10

The solution set is $\{1, 5\}$ .

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QuestionSolve the proportion. Express numbers as integers or simplified fractions. 1x−3=x−34\frac{1}{x-3}=\frac{x-3}{4}x−31​=4x−3​The solution set is □\square□ \}.

Studdy Solution

STEP 1

STEP 2

1. Cross-multiply2. Expand and rearrange3. Solve the quadratic equation

STEP 3

STEP 4

This simplifies to: 4=(x−3)24 = (x-3)^24=(x−3)2

STEP 5

Now, let's **expand** the right side of the equation: 4=x2−6x+94 = x^2 - 6x + 94=x2−6x+9

STEP 6

To **solve** this quadratic equation, we want to set it equal to zero.Let's subtract 4 from both sides: 0=x2−6x+50 = x^2 - 6x + 50=x2−6x+5

STEP 7

We can **factor** this quadratic equation!We are looking for two numbers that multiply to 5 and add up to -6.Those numbers are -5 and -1.So, we can rewrite the equation as: (x−5)(x−1)=0(x-5)(x-1) = 0(x−5)(x−1)=0

STEP 8

Now, we use the **zero product property**: if the product of two factors is zero, then at least one of the factors must be zero.This gives us two possible solutions: x−5=0orx−1=0x-5 = 0 \quad \text{or} \quad x-1 = 0x−5=0orx−1=0

STEP 9

STEP 10

The solution set is {1,5}\{1, 5\}{1,5}.

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QuestionSolve the proportion. Express numbers as integers or simplified fractions. $\frac{1}{x-3}=\frac{x-3}{4}$
The solution set is $\square$ \}.

1. Cross-multiply
2. Expand and rearrange
3. Solve the quadratic equation

This simplifies to: $4 = (x-3)^2$

Now, let's expand the right side of the equation: $4 = x^2 - 6x + 9$

To solve this quadratic equation, we want to set it equal to zero.
Let's subtract 4 from both sides: $0 = x^2 - 6x + 5$

We can factor this quadratic equation!
We are looking for two numbers that multiply to 5 and add up to -6.
Those numbers are -5 and -1.
So, we can rewrite the equation as: $(x-5)(x-1) = 0$

Now, we use the zero product property: if the product of two factors is zero, then at least one of the factors must be zero.
This gives us two possible solutions: $x-5 = 0 \quad \text{or} \quad x-1 = 0$

The solution set is $\{1, 5\}$ .