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trasher [3.6K]
3 years ago
6

Solve (4x – 3) > 5 for x.

Mathematics
2 answers:
kap26 [50]3 years ago
7 0

\huge\text{Hey there!}

\mathsf{(4x -  3) > 5}

\mathsf{4x - 3 > 5}

\large\text{ADD 3 to BOTH SIDES}

\mathsf{4x - 3 + 3 > 5 + 3}

\large\text{CANCEL out: -3 + 3 because that gives you 0}

\large\text{KEEP: 5 + 3 because it helps solve for x}

\mathsf{5 + 3 = \bf 8}

\large\text{New EQUATION:  }\mathsf{4x > 8}

\large\text{DIVIDE 4 to BOTH SIDES}

\mathsf{\dfrac{4x}{4}>\dfrac{8}{4}}

\large\text{CANCEL out: } \mathsf{\dfrac{4}{4}}\large\text{ because that gives you 1}

\large\text{KEEP: }\mathsf{\dfrac{8}{4}}\large\text{ because it helps you compares to x}

\mathsf{\dfrac{8}{4}=\bf 2}

\boxed{\boxed{\large\textsf{Therefore your ANSWER: \boxed{\mathsf{\bf x > 2}}}}}\huge\checkmark

\large\text{Good luck on your assignment and enjoy your day!}

~\frak{Amphitrite1040:)}

dmitriy555 [2]3 years ago
4 0

Answer:

x>2

Step-by-step explanation:

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GrogVix [38]

Answer:

B .75

Step-by-step explanation:

hope this helped <3

4 0
2 years ago
What is the sum of the terms of the series 4+7+10+...+22
skad [1K]
My answer is 13+16+19 because we’re are adding 3 each time
7 0
3 years ago
I need help please ASAP!! I'm in the middle of a quiz!
bixtya [17]

Answer:

2 is the correct answer

4 0
3 years ago
14. Write a two-column proof.
scZoUnD [109]

Given:

\bold{ 8x=4y-20} \\\\ \bold{ y=13}

Prove:

\bold{x=4}

Solution:

\to \bold{8x=4y-20}..........(i)\\\\\to \bold{ y=13}........................(ii)

Putting the value of equation (ii) into the equation (i):

\to \bold{8x=4(13)-20}\\\\\to \bold{8x=52-20}\\\\\to \bold{8x=32}\\\\\to \bold{x=\frac{32}{8}}\\\\\to \bold{x= 4}\\\\

putting the value x into the equation (i):

\to \bold{8(4)=4y-20}\\\\\to \bold{32=4y-20}\\\\\to \bold{32+20=4y}\\\\\to \bold{52=4y}\\\\\to \bold{y=\frac{54}{4}}\\\\\to \bold{ y= 13.5}

Therefore, the value of x= 4

Learn more:

brainly.com/question/6219408

3 0
2 years ago
4r – 4s + 4t=-4
Ber [7]

Answer:

1,3,1

Step-by-step explanation:

3 0
3 years ago
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