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By Simultaneous Equations. Assuming you mean
5x + 3y + 20 = 0
2x - y - 14 = 0.
We will eliminate 'y' by multiplying the bottom eq'n by '3' and then adding.
Hence
5x + 3y + 20 = 0
6x - 3y -42 = 0
Add
11x -22 = 0
11x = 22
x = 22/11
x = 2
when x= 2
2(2) - y - 14 =0
4 - y - 14 = 0
-y - 10 = 0
y = -10
So the point of intersection in ( x.y.) form is ( 2, -10).
What else can I help you with?
-2x - 14 equals 26?
- 2x - 14 = 26 Switch all signs: 2x + 14 = - 26 Subtract 14 from both sides: 2x = -26 - 14 = - 40 Divide both sides by 2: x = - 40/2 = - 20
3x plus 6y equals 18 2x plus 5y equals 20?
You need to be more clear about your question. 3x + 6y = 18 2x + 5y = 20 The two equations you give represent lines. I assume you're trying to find the point at which they intersect. There are various ways to work this out. One technique is to multiply each equation by a factor that gives one of the two variables the same coefficient. For example, if we multiply the first equation by 2, and the second equation by 3, we get: 6x + 12y = 36 6x + 15y = 60 Then we can subtract one from the other, which gives us a solution for y: [6x + 12y = 36] - [6x + 15y = 60] _______________ [0x - 3y = -24] So -3y = -24, telling us that at the point of intersection, y = 8. We can then plug that back into one of the original equations and solve for x: 2x + 5y = 20 2x + 5(8) = 20 2x = 20 - 40 2x = -20 x = -10 So the lines intersect at the point (-10, 8). Another way to work it out is to solve one of the equations for a single variable, and then plug it into the other one: 3x + 6y = 18 ∴ x + 2y = 6 ∴ x = 6 - 2y 2x + 5y = 20 ∴ 2(6 - 2y) + 5y = 20 ∴ 12 - 4y + 5y = 20 ∴ y = 20 - 12 ∴ y = 8 That gives us the value 8 for y, which we can once again plug in to either of the original equations to find x: 2x + 5y = 20 ∴ 2x + 40 = 20 ∴ 2x = -20 ∴ x = -10 Once again telling us that the lines intersect at the point (-10, 8)
What type of lines are these y 2x and y 2x -1?
As stated these are not lines, but just a collection of algebraic terms. If we change them to y=2x and y=2x-1, then on a graph of y versus x, these are parallel lines separated by vertical distance of 1.
2X-5 equals 15 FIND X?
2x - 5 = 15 2x = 20 x = 10
In rhombus ABDE diagonals AD and BE intersect at F. If AF 2x 7 and AD6x-8 find the value of x?
I think that the question is not clear please check it . = In rhombus ABDE diagonals AD and BE intersect at F. If AF 2x 7 and AD6x-8 find the value of x? = Have a look at the stricken out part of the question.
Solve -2x plus 14 equals 20?
-2x+14=20 -2x=20-14 -2x=6 -2x/-2=6/-2 x=-3
What is the answer to -2x plus 14 equals 20?
-2x+14=20 x=-3
What is -2x-6 equals 14?
-2x-6 = 14 -2x = 14+6 -2x = 20 x = -10
-2x - 14 equals 26?
- 2x - 14 = 26 Switch all signs: 2x + 14 = - 26 Subtract 14 from both sides: 2x = -26 - 14 = - 40 Divide both sides by 2: x = - 40/2 = - 20
Where do lines 6x-2y equals 4 and 2x plus y equals 8 intersect?
At the point (2, 4).
3x plus 6y equals 18 2x plus 5y equals 20?
You need to be more clear about your question. 3x + 6y = 18 2x + 5y = 20 The two equations you give represent lines. I assume you're trying to find the point at which they intersect. There are various ways to work this out. One technique is to multiply each equation by a factor that gives one of the two variables the same coefficient. For example, if we multiply the first equation by 2, and the second equation by 3, we get: 6x + 12y = 36 6x + 15y = 60 Then we can subtract one from the other, which gives us a solution for y: [6x + 12y = 36] - [6x + 15y = 60] _______________ [0x - 3y = -24] So -3y = -24, telling us that at the point of intersection, y = 8. We can then plug that back into one of the original equations and solve for x: 2x + 5y = 20 2x + 5(8) = 20 2x = 20 - 40 2x = -20 x = -10 So the lines intersect at the point (-10, 8). Another way to work it out is to solve one of the equations for a single variable, and then plug it into the other one: 3x + 6y = 18 ∴ x + 2y = 6 ∴ x = 6 - 2y 2x + 5y = 20 ∴ 2(6 - 2y) + 5y = 20 ∴ 12 - 4y + 5y = 20 ∴ y = 20 - 12 ∴ y = 8 That gives us the value 8 for y, which we can once again plug in to either of the original equations to find x: 2x + 5y = 20 ∴ 2x + 40 = 20 ∴ 2x = -20 ∴ x = -10 Once again telling us that the lines intersect at the point (-10, 8)
How do you find the equation of a line through 10 14 parallel to 1y equals -2x plus 4?
The slope will be the same but the y intercept will change:- y -14 = - 2(x - 10) y -14 = -2x +20 y = -2x +20 +14 y = -2x +34
-2 x - 14 equals 26?
-2x - 14 = 26 -2x = 26 + 14 x = -40/2 x = -20
What are two lines that have a point in common?
Two lines that have a point in common are said to be intersecting lines. When they intersect, they share a specific coordinate point where they cross each other. For example, the lines described by the equations y = 2x + 1 and y = -x + 3 intersect at the point (2, 5). This common point is where both lines meet on a graph.
Y equals 2x plus 2 and 4y-8x equals 12 at which point do these lines intersect?
They don't. The line is parallel. ( 4x-8x=12 simplifies to y-2x+3, which has the same slope as y=2x+2)
The solution for two lines that are parallel?
There is no solution. The answer to this sort of question (such as y=2x+3, and y=2x+4) would be no solution, since the lines never intersect, but instead continue on to go an infinite distance without ever crossing each other.
The system of equations 3x minus 6y equals 20 and 2x minus 4y equals 3 is?
The system of equations 3x - 6y = 20 and 2x - 4y = 3 is inconsistent. This is because the second equation can be derived from the first by multiplying by a factor, but the constants on the right side do not match, indicating that the lines represented by these equations are parallel and do not intersect. Therefore, there is no solution to the system.
