7x+4y = 39 2x+4y = 14 Subtract the equations in order to eliminate y: 5x = 25 Divide both sides by 5 to find the value of x: x = 5 Substitute the value of x into the original equations to find the value of y: Therefore: x = 5 and y = 1
7x + 5y 7(3) + 5(2) 21 + 10 = 31
(5, 2)
If: x -y = 2 then x^2 = y^2 +4y +4 If: x^2 -4y^2 = 5 then x^2 = 4y^2 +5 So: 4y^2 +5 = y^2 +4y +4 Transposing terms: 3y^2 -4y +1 = 0 Factorizing the above: (3y -1)(y -1) = 0 meaning y = 1/3 or y = 1 Substitution into the original linear equation intersections are at: (7/3, 1/3) and (3, 1)
2log4(7x) - 5 = 0 2log4(7x) = 5 log4(7x) = 5/2 7x = 45/2 =(41/2)5 = 25 = 32 x = 32/7
4y - 5 = 7x ∴y = 7/4x + 5/4 slope = 7/4, or 1.75 4y - 5 = 7x ∴4y/7 - 5/7 = x ∴4y/7 - 5/7 = 0 ∴4y = 5 ∴y = 1.2 the y-intercept is 1.2
7x+4y = 39 2x+4y = 14 Subtract the equations in order to eliminate y: 5x = 25 Divide both sides by 5 to find the value of x: x = 5 Substitute the value of x into the original equations to find the value of y: Therefore: x = 5 and y = 1
-7x + 5 = -9 -7x = -14 7x = 14 x = 2
If you mean: 4x -4y = 12 and -7x +4y = -36 then x = 8 and y = 5
8x plus 4y equals 5 is 8x + 4y = 5.
7x is still the slope, no matter what x equals.
23 = 7x + 7x - 5 = 14x - 5 14x = 23 + 5 = 28 x = 28/14 = 2
7x + 5y 7(3) + 5(2) 21 + 10 = 31
x - 2 = 2 → x = 4 → x² - 4y² = 5 → 4² - 4y² = 5 → 4y² = 16 - 5 → 4y² = 11 → y² = 11/4 → y = ± √(11/4) → The points of intersection of x - 2 = 2 with x² - 4y² = 5 are (4, -√(11/4)) ≈ 4, -1.658) and (4, √(11/4)) ≈ 4, 1.658)
(5, 2)
6x = 32 - 4y and 4y - 8 = 6x so 32 - 4y = 4y - 8 ie 40 = 8y so y = 5 and x = 2
yes.