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I presume that you mean 5x + 3y + 20 = 0 and 2x - y - 14 = 0.
First put the formulae in the form y = mx + c.
3y = -5x -20
y = -5/3 x -20/3
and
y = -2x + 14
Substitute the second formula into the first
-2x - 14 = -5/3 x -20/3
-6x - 42 = -5x - 20
-42 = x - 20
x = -22
Substitute into equation
y = -2x + 14
y = -2 x -22 + 14
y = 44 + 14
y = 58
Therefore the two lines meet at point (-22, 58).
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-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
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.
Which ordered pair is a soultion of the linear system 2x plus y equals 10 and 3y equals 2x plus 6?
x = 3 and y = 4 so the lines intersect at (3, 4)
How do you solve 3X plus 6 equals x plus 20?
balance the equation 3x-x=20-6 3x-x=2x 20-6=14 2x=14 x=7
