QuestionFind $x$ if $R S = x + 8$ , $S T = 2x - 6$ , and $R T = 32$ . Calculate $R S$ and $S T$ .

Studdy Solution

STEP 1

Assumptions1. The length of segment RS is $x+8$ . The length of segment is $x-6$
3. The length of segment RT is324. The segments RS,, and RT form a straight line, so $RS + = RT$

STEP 2

First, we need to set up an equation based on the information given. Since RS,, and RT form a straight line, we can say that the sum of RS and is equal to RT.
$RS + = RT$

STEP 3

Now, plug in the given values for RS,, and RT to form the equation.
$(x+8) + (2x-6) =32$

STEP 4

implify the left side of the equation by combining like terms.
$3x +2 =32$

STEP 5

Subtract2 from both sides of the equation to isolate the term with x.
$3x =30$

STEP 6

Divide both sides of the equation by3 to solve for x.
$x =10$

STEP 7

Now that we have the value of x, we can find the lengths of RS and by substituting x =10 into the expressions for RS and.
For RS $RS = x +$

STEP 8

Substitute x =10 into the expression for RS.
$RS =10 +8$

STEP 9

Calculate the length of RS.
$RS =18$

STEP 10

For $=2x -6$

STEP 11

Substitute x =10 into the expression for.
$=(10) -6$

STEP 12

Calculate the length of.
$=14$ The value of $x$ is10, the length of $RS$ is18, and the length of $$ is14.

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QuestionFind $x$ if $R S = x + 8$ , $S T = 2x - 6$ , and $R T = 32$ . Calculate $R S$ and $S T$ .

Studdy Solution

STEP 1

Assumptions1. The length of segment RS is $x+8$ . The length of segment is $x-6$
3. The length of segment RT is324. The segments RS,, and RT form a straight line, so $RS + = RT$

STEP 2

First, we need to set up an equation based on the information given. Since RS,, and RT form a straight line, we can say that the sum of RS and is equal to RT.
$RS + = RT$

STEP 3

Now, plug in the given values for RS,, and RT to form the equation.
$(x+8) + (2x-6) =32$

STEP 4

implify the left side of the equation by combining like terms.
$3x +2 =32$

STEP 5

Subtract2 from both sides of the equation to isolate the term with x.
$3x =30$

STEP 6

Divide both sides of the equation by3 to solve for x.
$x =10$

STEP 7

Now that we have the value of x, we can find the lengths of RS and by substituting x =10 into the expressions for RS and.
For RS $RS = x +$

STEP 8

Substitute x =10 into the expression for RS.
$RS =10 +8$

STEP 9

Calculate the length of RS.
$RS =18$

STEP 10

For $=2x -6$

STEP 11

Substitute x =10 into the expression for.
$=(10) -6$

STEP 12

Calculate the length of.
$=14$ The value of $x$ is10, the length of $RS$ is18, and the length of $$ is14.

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QuestionFind xxx if RS=x+8R S = x + 8RS=x+8, ST=2x−6S T = 2x - 6ST=2x−6, and RT=32R T = 32RT=32. Calculate RSR SRS and STS TST.

Studdy Solution

STEP 1

Assumptions1. The length of segment RS is x+8x+8x+8 . The length of segment is x−6x-6x−63. The length of segment RT is324. The segments RS,, and RT form a straight line, so RS+=RTRS + = RTRS+=RT

STEP 2

First, we need to set up an equation based on the information given. Since RS,, and RT form a straight line, we can say that the sum of RS and is equal to RT.RS+=RTRS + = RTRS+=RT

STEP 3

Now, plug in the given values for RS,, and RT to form the equation.(x+8)+(2x−6)=32(x+8) + (2x-6) =32(x+8)+(2x−6)=32

STEP 4

implify the left side of the equation by combining like terms.3x+2=323x +2 =323x+2=32

STEP 5

Subtract2 from both sides of the equation to isolate the term with x.3x=303x =303x=30

STEP 6

Divide both sides of the equation by3 to solve for x.x=10x =10x=10

STEP 7

Now that we have the value of x, we can find the lengths of RS and by substituting x =10 into the expressions for RS and.For RSRS=x+RS = x +RS=x+

STEP 8

Substitute x =10 into the expression for RS.RS=10+8RS =10 +8RS=10+8

STEP 9

Calculate the length of RS.RS=18RS =18RS=18

STEP 10

For=2x−6 =2x -6=2x−6

STEP 11

Substitute x =10 into the expression for.=(10)−6 =(10) -6=(10)−6

STEP 12

Calculate the length of.=14 =14=14The value of xxx is10, the length of RSRSRS is18, and the length of $$ is14.

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QuestionFind $x$ if $R S = x + 8$ , $S T = 2x - 6$ , and $R T = 32$ . Calculate $R S$ and $S T$ .

Assumptions1. The length of segment RS is $x+8$ . The length of segment is $x-6$
3. The length of segment RT is324. The segments RS,, and RT form a straight line, so $RS + = RT$

First, we need to set up an equation based on the information given. Since RS,, and RT form a straight line, we can say that the sum of RS and is equal to RT.
$RS + = RT$

Now, plug in the given values for RS,, and RT to form the equation.
$(x+8) + (2x-6) =32$

implify the left side of the equation by combining like terms.
$3x +2 =32$

Subtract2 from both sides of the equation to isolate the term with x.
$3x =30$

Divide both sides of the equation by3 to solve for x.
$x =10$

Now that we have the value of x, we can find the lengths of RS and by substituting x =10 into the expressions for RS and.
For RS $RS = x +$

Substitute x =10 into the expression for RS.
$RS =10 +8$

Calculate the length of RS.
$RS =18$

For $=2x -6$

Substitute x =10 into the expression for.
$=(10) -6$

Calculate the length of.
$=14$ The value of $x$ is10, the length of $RS$ is18, and the length of $$ is14.