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Five times a number decreased by nine is equal to twice the number increased by 23. Which equation could be used to solve the problem?

Rachna has nickels, dimes, and quarters in her piggy bank. She has $4.65 altogether. She has three more quarters than dimes and twice as many nickels as quarters. How many quarters does Rachna have?

Three times a number, x, increased by four is equal to five times the number, x. Which equation can be used to solve for x?

What is the value of x in the equation 0.2(x + 1) + 0.5x = –0.3(x – 4)?

Respuesta :

soon2bmommy
The first one is 5x-9=2x+23

The third one is 3x+4=5x
calculista

Answer:

Part 1) [tex]5x-9=2x+23[/tex]

Part 2) [tex]11\ quarters[/tex]

Part 3) [tex]3x+4=5x[/tex]

Part 4) [tex]x=1[/tex]

Step-by-step explanation:

Part 1)

Let

x-----> the number

The expression" Five times a number decreased by nine is equal to twice the number increased by [tex]23[/tex]", is equal to the algebraic expression

[tex]5x-9=2x+23[/tex]

Part 2) we have that

[tex]1\ nickel=\$0.05[/tex]

[tex]1\ dime=\$0.10[/tex]

[tex]1\ quarter=\$0.25[/tex]

Let

x------> the number of nickels

y-------> the number of dimes

z-------> the number of quarters

[tex]0.05x+0.10y+0.25z=4.65[/tex]

Multiply by [tex]100[/tex] both sides

[tex]5x+10y+25z=465[/tex] ------> equation A

[tex]z=3+y[/tex] -----> isolate the variable y

[tex]y=z-3[/tex] ------> equation B

[tex]x=2z[/tex] ------> equation C

substitute equation B and equation C in equation A

[tex]5(2z)+10(z-3)+25z=465[/tex]

[tex]10z+10z-30+25z=465[/tex]

[tex]45z=465+30[/tex]

[tex]45z=495[/tex]

[tex]z=11\ quarters[/tex]

Part 3) we know that

Let

x------> the number

The equation that represent the expression is equal to

[tex]3x+4=5x[/tex]

Part 4) we have

[tex]0.2(x+1)+0.5x=-0.3(x-4)[/tex]

Multiply by [tex]10[/tex] both sides

[tex]2(x+1)+5x=-3(x-4)[/tex]

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

[tex]7x+3x=12-2[/tex]

[tex]10x=10[/tex]

[tex]x=1[/tex]

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