All categories

2 years ago

Can anyone help me solve these linear systems using substitution?

Mathematics

2 answers:

Stolb23 [73]2 years ago

7 0

Answer:

The solutions to the sets of equations may be found by either a substitution process or by graphing and looking for the point that the two lines formed by the quations intersect. Both approaches are described.

The "solution" to these sets of equations is the point at which the two lines intersect each other.

Step-by-step explanation:

1. 3x-y=4 and x+2y=6

x+2y=6

6x-2y=8 We can simply add these two equations since y will disappear, leaving just x.

7x = 14 [result of addition]

x = 2 [Solve]

Since x = 2, [Then use the solution for x to find y]

x+2y=6

(2)+2y=6

2y = 4

y = 2

(2,2) is the solution (where the lines intersect). See attached graph.

2. 2x-y= -39 and x+y= -21

2x-y= -39

x+y= -21 Add the two equations (since the y will be eliminated)

3x = -60

x = -20

Find y:

x+y= -21

(-20)+y= -21

y = -1

Solution is (-20,-1)

3. 2x+y =11 and 6x-5y =9

2x+y =11

-6x+-3y = -33 [Multiply by -3 and then add to other equation. This allows us to eliminate x]

6x - 5y = 9

-8y = - 24

y = 3

2x+y =11

2x+(3) =11

2x = 8

x = 4

Solution is (4,3)

==================

See attached graph for all the graphed solutions.

==================

The substitution process also works if we isolate one of the 2 variables in one equation and then use the resulting value in the second eqaution. For example, in problem 1:

3x-y=4

x+2y=6

Pick either equation and isolate the x or y. I'll pick the second equation and isolate for x:

x = 6-2y

Now use this value of x (6-2y) in the other equation:

3x-y=4

3(6-2y)-y=4

18 - 6y - y = 4

-7y = -14

y = 2

Then use y=2 to fiond x, using either equation:

x+2y=6

x+2(2)=6

x = 2

The solution is (2,2).

This same approach can be used on all these problems. It avoids trying to visualize what can be done to an entire equation to eliminate one of the variables, but it takes a bit more paper.

All solutions are graphed in the attachment.