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ALGEBRA II // MODULE 1
FUNCTIONS AND INVERSES - 1.2
1.2
READY, SET, GO!
Name
Period
Date
READY
Topic: Solving for a variable
Solve for x.
1.
17 = 5x + 2
2. 2x² – 5 = 3x2 – 12x +
31
3. 11 = 2x + 1


Sagot :

Answer:

1. x = 3

2.x = 6

3. x = 5

I hope this helps you :D

To solve for variables, we start by isolating the variable term and then solve for the variable. The values of x in the given equations are 3, 6 and 5, respectively.

Equation (1)

[tex]17 = 5x + 2[/tex]

Collect like terms

[tex]5x = 17 -2[/tex]

[tex]5x = 15[/tex]

Divide both sides by 5

[tex]x =3[/tex]

Equation (2)

[tex]2x^2 - 5 = 3x^2 - 12x + 31[/tex]

Collect like terms

[tex]3x^2 - 2x^2 - 12x + 31 +5=0[/tex]

[tex]x^2 - 12x + 36=0[/tex]

Expand

[tex]x^2 - 6x - 6x + 36=0[/tex]

Factorize

[tex]x(x - 6) - 6(x -6)=0[/tex]

Factor out x - 6

[tex](x - 6)(x -6)=0[/tex]

This means that:

[tex]x -6=0[/tex]

Solve for x

[tex]x = 6[/tex]

Equation (3):

[tex]11 = 2x + 1[/tex]

Collect like terms

[tex]2x = 11-1[/tex]

[tex]2x = 10[/tex]

Divide both sides by 2

[tex]x =5[/tex]

Hence, the values of x in the given equations are 3, 6 and 5.

Read more about equations at:

https://brainly.com/question/953809

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