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What are the solutions to the simultaneous equations of 3x -2y equals 1 and 3x squared -2y squared plus 5 equals 0 showing work 1?
1st equation: 3x-2y = 1 or 3x = 1+2y 2nd equation: 3x^2 -2y^2 +5 = 0 or 3x^2 = 2y^2 -5 Square both sides in 1st equation and multiply all terms in 2nd equation by 3 If: 9x^2 = 1+4y+4y^2 and 9x^2 = 6y^2 -15 Then: 6y^2 -15 = 1+4y+4y^2 => 2y^2 -4y-16 = 0 Dividing all terms by two: y^2 -2y-8 = 0 Factorizing: (y-4)(y+2) = 0 => y = 4 or y = -2 Solutions by substitution: x = 3 and y = 4 or x = -1 and y = -2
How do you write this equation 3x equals 2y plus 3 in function form?
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
What is the graph of the equation -3x plus 4y equal 12?
-3x + 4y = 12∴ 4y = 3x + 12∴ y = 3x/4 + 3So this is a line with a slope of 3/4, an x-intercept of {0, 3} and a y-intercept of {-4, 0}
Solve -3x plus 4y equals 12?
-3x + 4y = 12 Temporary value of y: 4y = 3x + 12 y = (3x + 12) / 4 Substituting y: -3x + 4 [(3x+12) / 4] = 12 -3x + 12x + 48 = 12 -3x + 12x = 12 - 48 9x = 12 - 48 9x = -36 x = -36 / 9 x = -4 Subtituting x to get the final value of y: -3(-4) + 4y = 12 12 + 4y = 12 4y = 12 - 12 4y = 0 y = 0 Final answer: x = -4 y = 0 Check: -3x + 4y = 12 -3(-4) + 4(0) = 12 12 + 0 = 12 12 = 12 Solution 2: -3x + 4y = 12 Temporay value of x: 3x = 4y - 12 x = (4y - 12) / 3 Substituting x to get the value of y: 3[(4y - 12) / 3] + 4y = 12 12y - 36 + 4y = 12 16y = 12 + 36 16y = 48 y = 48 / 16 y = 3 Substituting y to get the value of x: -3x + 4(3) = 12 -3x + 12 = 12 -3x = 12 - 12 x = 0 Final Answer: x = 0 y = 3 Check: -3(0) + 4(3) = 12 12 = 12
Y intercept of 2y equals -6x plus 8?
2y = -6x + 8 The y intercept occurs when x = 0 2y = 8 therefore y = 4
What type of lines are these 3x - 2y -2 and 6x - 4y 0?
If you mean: 3x-2y = -2 and 6x-4y = 0 then they are parallel lines having the same slope or gradient of 1.5
What type of lines are these 3x-2y equals -2 and 6x-4y equals 0?
parallel
What type of lines are these 3x - 2y equals -2 and 6x - 4y equals 0?
They are parallel straight lines,
What type of lines are these 3x-2y equals -2 and 6x-4y-0?
They are straight lines. They are positively inclined. They have the same gradient.
Which type of lines are these 3x-2y equals -2 and 6x-4y equals 0?
3x - 2y = -2 and 6x - 4y = 0Solving both equations for y, we havey = (3/2)x + 1 and y = (3/2)xSince both lines have the same slope, they are parallel lines.
What is the answer of 6x plus 2y-3 equals 0?
6x+2y-3 = 0 2y = -6x+3 y = -3x+3/2 in slope intercept form
What is the x-intercept of 3x-2y equals 6?
3x - 2y = 6At the x-intercept, y=0 :3x = 6x = 2
What are the solutions to the simultaneous equations of 3x -2y equals 1 and 3x squared -2y squared plus 5 equals 0 showing work 1?
1st equation: 3x-2y = 1 or 3x = 1+2y 2nd equation: 3x^2 -2y^2 +5 = 0 or 3x^2 = 2y^2 -5 Square both sides in 1st equation and multiply all terms in 2nd equation by 3 If: 9x^2 = 1+4y+4y^2 and 9x^2 = 6y^2 -15 Then: 6y^2 -15 = 1+4y+4y^2 => 2y^2 -4y-16 = 0 Dividing all terms by two: y^2 -2y-8 = 0 Factorizing: (y-4)(y+2) = 0 => y = 4 or y = -2 Solutions by substitution: x = 3 and y = 4 or x = -1 and y = -2
What is the equation of the tangent line that touches the circle x squared plus y squared -6x plus 4y equals 0 at the point of 6 -4 on the Cartesian plane?
Equation: x² + y² -6x +4y = 0 Completing the squares: (x-3)² + (y+2)² = 13 Centre of circle: (3, -2) Contact point: (6, -4) Slope of radius: -2/3 Slope of tangent: 3/2 Tangent equation: y - -4 = 3/2(x-6) => 2y - -8 = 3x-18 => 2y = 3x-26 Tangent line equation in its general form: 3x-2y-26 = 0
3x y - 2y?
3xy-2y=0 3xy=2y y=2y (3x) y/2y=3x 1/2=3x multiply across by 2 1=6x 1/6=x therefore substituting x=1/6 into 3xy-2y; 3(1/6)y-2y=0 1/2y=2y y=2y/0.5 0.5 aka 1/2 y=1
What is the point of contact when the tangent line 3x -2y -26 equals 0 touches the circle x squared plus y squared -6x plus 4y equals 0 on the Cartesian plane showing work?
If: 3x -2y -26 = 0 Then: y = 1.5x -13 If: x^2 +y^2 -6x +4y = 0 Then: x^3 +(1.5x -13)^2 -6x +4(1.5x -13 = 0 Expanding brackets: x^2 +2.25x^2 -39x +169 -6x +6x -52 = 0 Collecting like terms: 3.25x^2 -39x +117 = 0 Divide all terms by 3.25: x^2 -12x +36 = 0 Factorizing: (x-6)(x-6) = 0 => x = 6 and also x = 6 By substitution point of contact is at: (6, -4)
What are the solutions to the simultaneous equations of 3x -2y equals 1and 3x squared -2y squared plus 5 equals 0 showing work?
If: 3x-2y = 1 Then: 9x^2 = (1+2y)^2 by squaring both sides If: 3x^2 -2y^2 +5 = 0 Then: 9x^2 = 6y^2 -15 by multiplying all terms by 3 So: 6y^2 -15 = 1+4y +4y^2 Transposing and collecting like terms: 2y^2 -16 -4y = 0 Divide all terms by two: y^2 -8 -2y = 0 Factorizing: (y-4)(y+2) = 0 => y = 4 or y = -2 Therefore solutions by substitution are: (3, 4) and (-1, -2)
