WILL MARK BRAINLIEST. Can someone help me wit des 4 questions? SHOW UR WORK FOR EACH QUESTION
1.) 5(2x+6)=8x+48
2.) -3(8+3h)=5h+4
3.) 2(y-6)=3(y-4)-y
4.) 8x-4=2(4x-4)

Answers

Answer 1

Answer:

1. [tex]x[/tex] = 9

2. [tex]h[/tex] = 2

3. Infinitely Many Solutions

4. No solution

Step-by-step explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

1.) 5(2[tex]x[/tex] + 6) = 8[tex]x[/tex] + 48

Step 1: Expand.

10[tex]x[/tex] + 30 = 8

Step 2: Subtract 30 from both sides.

10[tex]x[/tex] = 8

Step 3: Simplify 8[tex]x[/tex] + 48 − 30 to 8[tex]x[/tex] + 18.

10[tex]x[/tex] = 8

Step 4: Subtract 8[tex]x[/tex] from both sides.

10[tex]x[/tex] − 8

Step 5: Simplify 10[tex]x[/tex] − 8[tex]x[/tex] to 2[tex]x[/tex].

2[tex]x[/tex] = 18

Step 6: Simplify [tex]\frac{18}{2}[/tex] to 9.

[tex]x[/tex] = 9

2.) −3(8 + 3[tex]h[/tex]) = 5[tex]h[/tex] + 4

Step 1: Expand.

−24 − 9[tex]h[/tex] = 5

Step 2: Add 24 to both sides.

−9[tex]h[/tex] = 5

Step 3: Simplify 5[tex]h[/tex] + 4 + 24 to 5[tex]h[/tex] + 28.

−9[tex]h[/tex] = 5

Step 4: Subtract 5[tex]h[/tex] from both sides.

−9[tex]h[/tex] − 5

Step 5: Simplify −9[tex]h[/tex] − 5[tex]h[/tex] to −14[tex]h[/tex].

−14[tex]h[/tex] = 28

Step 6: Divide both sides by −14.

[tex]h[/tex] = [tex]-\frac{28}{14}[/tex]

Step 7: Simplify [tex]\frac{28}{14}[/tex] to 2.

[tex]h[/tex] = 2

3.) 2([tex]y[/tex] − 6) = 3([tex]y[/tex] − 4) − [tex]y[/tex]

Step 1: Expand.

2[tex]y[/tex] − 12 = 3

Step 2: Simplify 3[tex]y[/tex] − 12 − [tex]y[/tex] to 2[tex]y[/tex] − 12.

2[tex]y[/tex] − 12 = 2 − 12

Step 3: Since both sides equal, there are infinitely many solutions.

Infinitely Many Solutions

4.) 8[tex]x[/tex] − 4 = 2(4[tex]x[/tex] − 4)

Step 1: Expand.

8[tex]x[/tex] − 4 = 8

Step 2: Cancel 8[tex]x[/tex] on both sides.

−4 = −8

Step 3: Since −4 = −8 − 4 = −8 is false, there is no solution.

No solution

~I hope I helped you! Good luck :)~


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Step-by-step explanation:

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Answers

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Answers

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Answers

Answer:

[tex] \overline{AC} [/tex] is congruent to [tex] \overline{DF} [/tex], because [tex] \overline{AC} = 9in [/tex], [tex] \overline{DF} = 8 in [/tex]. <C is congruent to <F. [tex] \overline{EF} [/tex] must correspond to [tex] \overline{BC} [/tex]. The triangles are congruent by the SAS Triangle

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Step-by-step explanation:

From the figures given, ∆ABC and ∆DEF have an included corresponding angles measuring 30° each (<C and <F), and also a given corresponding side length of 9 in each (side AC and side DF). The other given as am expression, and hence the length is unknown. For these two ∆s to be considered congruent, using the SAS Triangle Congruence Theorem, we must have two sides and an included angle of ∆ABC equal to the corresponding sides of and an included angle of ∆DEF.

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The answer is:

[tex] \overline{AC} [/tex] is congruent to [tex] \overline{DF} [/tex], because [tex] \overline{AC} = 9in [/tex], [tex] \overline{DF} = 8 in [/tex]. <C is congruent to <F. [tex] \overline{EF} [/tex] must correspond to [tex] \overline{BC} [/tex]. The triangles are congruent by the SAS Triangle

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Answers

[tex] \frac{1}{2} x[/tex]

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