Algebra 2 11th grade

4. If x = y + 3, then in the second equation you can write

-3 (y + 3) + 3y = 6

Expanding and simplifying gives

-3y - 9 + 3y = 6

-9 = 6

but this is clearly not true, so this system as no solution.

5. Solve the first equation for x :

x + 2y = 4   ===>   x = -2y + 4

Substitute into the other equation:

2 (-2y + 4) - 3y = 1

-4y + 8 - 3y = 1

-7y = -7

y = 1

Solve for x :

x = -2•1 + 4

x = -2 + 4

x = 2

6. Solve the first equation for y :

2x - y = 6   ===>   y = 2x - 6

Substitute into the other equation:

x + 2 (2x - 6) = -2

x + 4x - 12 = -2

5x = 10

x = 2

Solve for y :

y = 2•2 - 6

y = 4 - 6

y = -2

7. Solve either equation for x :

x + 4y = 10   ===>   x = -4y + 10

Substitute into the other equation:

(-4y + 10) - 2y = -8

-6y = -18

y = 3

Solve for x :

x = -4•3 + 10

x = -12 + 10

x = -2

8. Solve the first equation for y :

-2x + y = 1   ===>   y = 2x + 1

Substitute into the other equation:

4x - 2 (2x + 1) = 5

4x - 4x - 2 = 5

-2 = 5

This is false, so there is no solution.

9. Solve the second equation for x :

-x - 4y = -3   ===>   x = -4y + 3

Substitute into the other equation:

2 (-4y + 3) + 8y = 6

-8y + 6 + 8y = 6

0 = 0

This means any value of x and y will satisfy these two equations, so there are infinitely many solutions.