Solve by substitution. If there is not a unique solution, state whether the system has no solution or infinitely many solutions.

*y*
= 4*x* – 10

*y*
= 5 – *x*

Solve by substitution. If there is not a unique solution, state whether the system has no solution or infinitely many solutions.

2*x* + 2*y* = 0

6*x* + *y* = –10

Solve by substitution. If there is not a unique solution, state whether the system has no solution or infinitely many solutions.

2*x* + 3*y* = 4

*y* = 5*x* – 27

*c* – 3*d* = 2

3*c* + *d* = 16

*x* + 4*y* = 19

*x* – 2*y* = 1

3*r* – *s* = 3

–6*r* + 5*s* = 21

7*x* + 3*y* = 68

*x* – 4*y* = –8

–0.5*x* + *y* = 1.5

0.8*x* – 0.2*y* = 6.0

*y* = 4*x* – 6

3*x* – 5*y* = 12

*x* + *y*
= 9

*y* = 3*x*
+ 1

*x* = *y*
– 2

*y* = 10 – 3*x*

*y* = 3*x*
– 7

6*y* – *x*
= 9

*x* = –4*y*

* x* = 4 – 6*y*

*x* = 5*y*
– 2

3*x* – *y*
= 8

*x* – *y*
= 9

*x* + *y*
= –3

*x* – *y*
= 3

*x* + 4*y*
= 7

*x* + 3*y*
= 25

4*x* + 5*y*
= 9

4*b* + 3*a*
= 5

–3*b* + *a*
= 6

*y* – 3*x*
= 0

2*x* + 6*y*
= 60

6*x* + 5*y*
= 4

3*x* = 2 – *y*

By substitution, solve the system of equations. Write the solution as an ordered triple of the form (*x*, *y*, *z*).

15*x*
– *y* + 6*z* = 108

*x*
+ *z* = 5

*y*
+ 4*z* = 2

Solve the system of equations. Write the solution as an ordered triple of the form (*x*,*y*, *z*).

*x*
+ *y* + *z* = –25

*y*
= –8*z*

*x*
= 12*z*

Car A costs $10,000, and costs $0.12 per mile to maintain. Car B costs $11,000 and costs $0.11 per mile to maintain. Suppose both cars are driven the same number of miles. At what mileage would the total costs of the two cars be the same?