Points: (6, -3) and (-4, -9) Slope: 3/5 Equation: 5y = 3x-33
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The value of X in the equation 3x 2-12x plus 9 is 1.25.
The problem '3x+9=81-5' can be solved as follows - 3x+9=81-5x Original Problem 3x=72-5x Subtract 9 from both sides of the equation 8x=72 Add '5x' to both sides of the equation x=9 Divide both sides by 8
3x+3=30 X=9
Points: (6, -3) and (-4, -9) Slope: 3/5 Equation: 5y = 3x-33
Points: (6, -3) and (-4, -9) Slope: 3/5 Equation: 5y = 3x-33
Without an equality sign the given terms can't be considered to be an equation.
40 + 9 = 5y + 9 Since 9 is present on both sides of the equation,therefore we can strike it off on both sides,the remaining equation being 40 = 5y or 5y = 40 5y = 40 or y = 40/5 Therefore,y = 9 .
3x + 9 y = 120 5x + 5y = 90 multiply first equation by 5 ( both sides) multiply second equation by 3 ( both sides) 15 x + 45 y = 600 15 x + 15 y = 270 subtract 30y = 330 divide by 30 y = 11 substiute y into first equation 3 x + 9 y = 120 3x + 99 = 120 3x = 21 x = 7
From first equation, -y = 3x + 3. Substitute in second equation: -3x + 5(3x + 3) = -21 ie 12x = -36 so x = -3 and y = -(-9 + 3) = 6. Easier method: subtract first equation from second giving -4y = -24 so y = 6, this in first equation gives -6 = 3x + 3, ie 3x = -9 so x = -3
It is 3x + 5y = 155 The slope of the line 5x -3y = 9: 5x - 3y = 9 → 3y = 5x - 9 → y = 5/3 x - 3 → slope is 5/3 → slope of perpendicular line is -3/5 Intercept at x = 15 → y = 5/3 x 15 - 3 = 22 → equation of perpendicular line is: y - 22 = -3/5 (x - 15) → 5y - 110 = -3x + 45 → 3x + 5y = 155
3x+5y=44 8x-7y=-25 Multiply the first equation by 7 and the second by 5. 21x+35y=308 40x-35y=-125 Put them together. 61x=183 x=3 Substitute this back into either equation. 3(3)+5y=44 9+5y=44 5y=35 y=7 To check put both into the other equation. 8(3)-7(7)=-25 24-49=-25 -25=-25 Yeah!
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3x-6y=9 -3x -3x -6y=9-3x __ ____ -6 -6 y=9-3x ____ -6
If you mean points of (-5, 0) and (10, 9) then the slope is 3/5 and the straight line equation is 5y = 3x+15
3x - 2 = 9 3x = 9 + 2 = 11 x = 11/3