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A system of equations is shown below:

n = 3m + 7
n โˆ’ 2m = 1

What is the solution, in the form (m, n), to the system of equations?

(3, 7)
(1, 8)
(โˆ’3, โˆ’2)
(โˆ’6, โˆ’11)


Which statement is true about the solutions for the equation 4y + 6 = โˆ’2?

It has infinitely many solutions.
It has two solutions.
It has one solution.
It has no solution.


The system of equations shown below is graphed on a coordinate grid:

3y + x = 4
2y โˆ’ x = 6

Which statement is true about the coordinates of the point that is the solution to the system of equations?

It is (โˆ’2, 2) and lies on both lines.
It is (โˆ’5, 3) and lies on both lines.
It is (โˆ’5, 3) and does not lie on either line.
It is (โˆ’2, 2) and does not lie on either line.


What is the value of z in the equation 2z + 6 = โˆ’4?

5
1
โˆ’1
โˆ’5


Which statement is true for the equation 2x โˆ’ 2x โˆ’ 7 = โˆ’7?

It has infinitely many solutions.
It has two solutions.
It has one solution.
It has no solution.


The incomplete work of a student to solve an equation is shown below:

Step 1: 4x + 12 = 4
Step 2: ?
Step 3: x = โˆ’8 รท 4
Step 4: x = โˆ’2

What is the missing Step 2?

4x = 8
4x = 16
4x = โˆ’16
4x = โˆ’8



Which set of steps shows the solution to the equation 2y = โˆ’8?

y = โˆ’8 โˆ’ 2; y = 6
y = โˆ’8 รท (โˆ’2); y = 4
y = โˆ’8 รท 2; y = โˆ’4
y = โˆ’8 โˆ’ (โˆ’2); y = โˆ’6



A system of two equations is shown below:

Equation C: a = 3b + 6
Equation D: a = 7b โˆ’ 1

What value of a can be substituted into equation D to solve the system of equations?

3b
7b
3b + 6
7b โˆ’ 1

Respuesta :

Answer:

1. (-6, -11)

2. It has one solution

3. It is (โˆ’2, 2) and lies on both lines

4. [tex]-5[/tex]

5. It has infinitely many solutions.

6. [tex]4x=-8[/tex]

7. y = โˆ’8 รท 2; y = โˆ’4

8. 3b + 6


Step-by-step explanation:


Question 1:

Substituting equation 1 into equation 2 and solving for [tex]m[/tex] gives us:

[tex]n-2m=1\\(3m+7)-2m=1\\m+7=1\\m=1-7\\m=-6[/tex]

Plugging this value into equation 1 gives us [tex]n[/tex], so we have:

[tex]n=3m+7\\n=3(-6)+7\\n=-18+7\\n=-11[/tex]

Hence, the solution, in the form (m, n), to the system of equations is (-6,-11).


Question 2:

Solving the equation for [tex]y[/tex] gives us:

[tex]4y+6=-2\\4y=-2-6\\4y=-8\\y=\frac{-8}{4}=-2[/tex]

As we can see, there is only one solution.


Question 3:

We can add the both equations so [tex]x[/tex] cancels out and then we can solve for [tex]y[/tex]:

[tex](3y+x=4)\\+(2y-x=6)\\---------\\5y=10\\y=\frac{10}{5}=2[/tex]

Substituting this value of [tex]y[/tex] into any equation above (let's use equation 1) will give us [tex]x[/tex]:

[tex]3y+x=4\\3(2)+x=4\\6+x=4\\x=4-6\\x=-2[/tex]

So the intersection point (or solution) (-2, 2) lies on both the lines.


Question 4:

Let's do some algebra and figure out the value of [tex]z[/tex]:

[tex]2z+6=-4\\2z=-4-6\\2z=-10\\z=\frac{-10}{2}=-5[/tex]

[tex]z[/tex] is -5


Question 5:

Reducing the equation gives us:

[tex]2x-2x-7=-7\\0-7=-7\\-7=-7[/tex]

We can plug in ANY VALUE into [tex]x[/tex] and make this equation true. So there are INFINITELY MANY SOLUTIONS.


Question 6:

Step 2 of the solution should be taking 12 to the other side so that variable is on one side and all the numbers to the other. So 2nd step would be:

[tex]4x+12=4\\4x=4-12\\4x=-8[/tex]

Rest of the steps follow. So, 2nd step would be [tex]4x=-8[/tex].


Question 7:

The next step to solving this equation would be to DIVIDE -8 by 2 since 2 is multiplied with [tex]y[/tex].

[tex]2y=-8\\y=\frac{-8}{2}=-4[/tex]

Third answer choice is right.


Question 8:

We can substitute the value of [tex]a[/tex] given in Equation C into Equation D to solve the system of equations.

The value of [tex]a[/tex] in Equation C is given as [tex]a=3b+6[/tex]

Third answer choice is right.