Solve the system of equations algebraically. 4x – 3y = 8 5x – 4y = 5 a (16, 21) b. many solutions c. no soluti

Question

Solve the system of equations algebraically.
4x – 3y = 8
5x – 4y = 5
a (16, 21)
b. many solutions
c. no solution
d. (17,20)

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Evelyn 3 weeks 2021-09-28T04:50:56+00:00 1 Answer 0

Answers ( )

    0
    2021-09-28T04:52:04+00:00

    Answer: D

    Step-by-step explanation:

    4x-3y=8\\5x-4y=5

    Let’s take one of the equations and solve for a variable to use the substitution method.

    5x-4y=5\\5x=5+4y\\x=\frac{5+4y}{5}

    Now replace this in the first equation.

    4x-3y=8\\4(\frac{5+4y}{5})-3y=8\\\frac{20+16y}{5}-3y=8

    Separate the terms.

    \frac{20}{5}+\frac{16y}{5}-3y=8

    Solve 20/5

    4+\frac{16y}{5}-3y=8

    Subtract 4 from both sides to isolate y.

    4-4+\frac{16y}{5}-3y=8-4

    \frac{16}{5}y-3y=4

    Solve the difference.

    \frac{16-(3)(5)}{5}y=4\\

    \frac{16-15}{5} y=4

    \frac{1}{5}y=4

    Multiply by the reciprocal or the inverted fraction that is next to y to isolate it.

    (\frac{5}{1} )(\frac{1}{5})y=4(\frac{5}{1})

    y=20

    Now, in order to find x, replace y in any of the two equations.

    5x-4y=5\\5x-4(20)=5\\5x-80=5\\5x=5+80\\5x=85\\x=\frac{85}{5}\\x=17

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45:7+7-4:2-5:5*4+35:2 =? ( )