Question

Solve the system of equations algebraically.

Answer 1

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.

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