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The "equation" has no visible signs.
It is a straight line equation that can be rearranged into slope intercept form: -x+2y = 12 2y = x+12 y = 0.5x+6 which is now in slope intercept form
Changing the equation from (4x + 2y = 12) to (4x + 2y = 36) will affect the position of the line on the graph. Specifically, the new equation represents a line that is parallel to the original line but shifted upwards. This is because both equations have the same slope, but the y-intercept changes, resulting in a higher intercept on the y-axis for the new equation. Thus, the line will intersect the y-axis at (y = 18) instead of (y = 6).
The slope (m) is -2.5 -2y = 5x-12 -2y/-2 = (5x-12)/-2 y=-2.5x + 6 or more neatly: y= 6-2.5x
Without an equality sign and not knowing the plus or minus values of 2y and 5 it can't be considered to be a straight line equation.
2x+y=-12-4x-2y=30 x=-y/2+30, you plug this into -12-4x-2y=30 and get -12+2y-120-2y=30 which gives 2y-2y=162. This is not possible, so the equation is unsolvable.
what is the slope of the line that has the equation 4x+2y=12?
3x + 2y = 8 This is an equation. It could be the equation of a line.
(0,6)
It is a straight line equation that can be rearranged into slope intercept form: -x+2y = 12 2y = x+12 y = 0.5x+6 which is now in slope intercept form
The slope (m) is -2.5 -2y = 5x-12 -2y/-2 = (5x-12)/-2 y=-2.5x + 6 or more neatly: y= 6-2.5x
2y= 3x+6
-2x - 2y = -122x + 2y = 122y = 12 - 2xy = 6 - x
It works out that the tangent line x -2y +12 = 0 makes contact with the circle x^2 +y^2 -x -31 = 0 at the point (-2, 5) ------------------------------------------------- The equation of the tangent line can be rearranged to isolate x: x - 2y + 12 = 0 → x = 2y - 12 This can now be substituted for x in the equation of the circle and solved to find the value(s) of y: x² + y² - x - 31 = 0 → (2y - 12)² + y² - (2y - 12) - 31 = 0 → 4y² - 48y + 144 + y² - 2y + 12 - 31 = 0 → 5y² - 50y - 125 = 0 → y² - 10y - 25 = 0 → (y - 5)² = 0 This is a repeated root showing a single point of contact, so the line IS a tangent to the circle. → y - 5 = 0 → y = 5 Substitute into the equation for the line: → x = 2y - 12 = 2×5 - 12 = 10 - 12 = -2 → point of contact is (-2, 5)
We solved the first equation for 'x': [ x = 2y + 8 ].Then we substituted it for 'x' in the second equation and rearranged: [ y2 + 4y - 12 = 0 ].Solutions of this quadratic equation are: y = 2 and -6. From which x = 12 and -4 .So the straight line intersects the hyperbola at (-4, -6) and at (12, 2) .
3x + 2y = 12 ie 2y = 12 - 3x so y = 6 - 3x/2
Impossible to say because the equal, the positive and/or the negative signs or instructions are missing in your question.
Without an equality sign the given expression can't be considered to be a straight line equation