Answer:
1. a. b = - 8
b. x = 8
c. x = 11
d. x = 5
2. 12 soccer balls and 8 basketballs can be purchased.
Step by step explanation
a. [tex] - 14 + 6b + 7 - 2b = 1 + 5b[/tex]
Calculate the sum
[tex] - 7 + 6b - 2b = 1 + 5b[/tex]
Collect like terms
[tex] 7 + 4b = 1 + 5b[/tex]
Move variable to L.H.S and change it's sign
Similarly, Move constant to R.H.S and change its sign
[tex]4b - 5b = 1 + 7[/tex]
Collect like terms
[tex] - b = 8[/tex]
Change the signs on both sides of the equation
[tex]b = - 8[/tex]
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b. [tex] \frac{5x + 10}{ - 6} = - 5[/tex]
Apply cross product property
[tex]5x + 10 = - 5 \times ( - 6)[/tex]
Multiply the numbers
[tex]5x + 10 = 30[/tex]
Move constant to R.H.S and change its sign
[tex]5x = 30 - 10[/tex]
Calculate the difference
[tex]5x = 20[/tex]
Divide both sides of the equation by 5
[tex] \frac{5x}{5} = \frac{20}{5} [/tex]
Calculate
[tex]x = 4[/tex]
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c. [tex] - 15 = \frac{ - 8x - 17}{7} [/tex]
Apply cross product property
[tex] - 15 \times 7 = - 8x - 17[/tex]
Multiply the numbers
[tex] - 105 = - 8x - 17[/tex]
Swap the sides of the equation
[tex] - 8x - 17 = - 105[/tex]
Move constant to R.H.S and change its sign
[tex] - 8x = - 105 + 17[/tex]
Calculate
[tex] - 8x = - 88[/tex]
Change the signs on both sides of the equation
[tex]8x = 88[/tex]
Divide both sides of the equation by 8
[tex] \frac{8x}{8} = \frac{88}{8} [/tex]
Calculate
[tex]x = 11[/tex]
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D. [tex]5 = 6x + 5(x - 10)[/tex]
Distribute 5 through the parentheses
[tex]5 = 6x + 5x - 50[/tex]
Collect like terms
[tex]5 = 11x - 50[/tex]
Swap both sides of the equation
[tex]11x - 50 = 5[/tex]
Move constant to R.H.S and change its sign
[tex]11x = 5 + 50[/tex]
Calculate the sum
[tex]11x = 55[/tex]
Divide both sides of the equation by 11
[tex] \frac{11x}{11} = \frac{55}{11} [/tex]
Calculate
[tex]x = 5[/tex]
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2.
Solution,
No.of students in soccer = x
No.of students in basketball = y
Total no.of students = 20
i.e x + y = 20 → equation ( i )
Cost of soccer ball = $ 7
Cost of basketball = $ 10
Total budget = $ 164
i.e 7x + 10 y = 165 → equation ( ii )
In equation ( i ),
x + y = 20
Move 'y' to R.H.S and change its sign
x = 20 - y
Put the value of x in equation ( i )
[tex]7(20 - y) + 10y = 164[/tex]
[tex]140 - 7y + 10y = 164[/tex]
[tex]3y = 164 - 140[/tex]
[tex]3y = 24[/tex]
[tex]y = \frac{24}{3} [/tex]
[tex]y = 8[/tex]
Now, put the value of y in equation ( i ) ,
x + y = 20
[tex]x + 8 = 20[/tex]
[tex]x = 20 - 8[/tex]
[tex]x = 12[/tex]
Hence, 12 soccer balls and 8 basketballs can be purchased.
Hope this helps...
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