7x + 51x + 14= 58x + 14= 2(29x + 7)
It's a third-degree polynomial in 'x'. It's value depends on the value of 'x'. Every time 'x' changes, the value of the polynomial changes.
My best guess is that you have an exercise which is headed (and thus gives the context): "solve for x in the following equations", though it could quite easily be: solve for x in the following inequalities".As such, there is some sign between the 26 and 8x which unfortunately has been stripped out; it is one of {=≤≥}If I solve it for equals (=), then the others will follow:-58x - 26 = 8x - 230.6→ -26 + 230.6 = 58x + 8x→ 204.6 = 64x→ 3.196875 = xI have deliberately left the x on the right so that when the inequalities are inserted, the correct answer is given:-58x - 26 = 8x - 230.6 → 3.196875 = x, or with x on the left: x = 3.196875-58x - 26 < 8x - 230.6 → 3.196875 < x, or with x on the left: x > 3.196875-58x - 26 > 8x - 230.6 → 3.196875 > x, or with x on the left: x < 3.196875-58x - 26 ≤ 8x - 230.6 → 3.196875 ≤ x, or with x on the left: x ≥ 3.196875-58x - 26 ≥ 8x - 230.6 → 3.196875 ≥ x, or with x on the left: x ≤ 3.196875Note the change of the inequalities when the x is shifted from the right to the left so that the same meaning is read.
If d is the highest common factor of 58 and 145 then d equals 29:- So: 29 = 58x+145y Divide all terms by 29: 1 = 2x+5y Or: 2*1/4+5*1/10 = 1 Therefore the possible values of x and y are 1/4 and 1/10 respectively Check: 58*1/4 plus 145*1/10 = 29
x = 4/58x + 4 = 3x + 8Subtract 3x from both sides5x + 4 = 8Subtract 4 from both sides5x = 4Divide both sides by 5x = 4/5Check Work:8( 4/5 ) + 4 = 3( 4/5 ) + 88/1 ∙ 4/5 + 4 = 3/1 ∙ 4/5 + 832/5 + 20/5 = 12/5 + 40/5Remove the common denominator32 + 20 = 12 + 4052 = 52
56x^2 + 96x + 40
15x2 + 58x + 48= 15x2 + 18x + 40x + 48= 3x(5x + 6) + 8(5x + 6)= (3x + 8)(5x + 6)
7x + 51x + 14= 58x + 14= 2(29x + 7)
2(4x - 1)(5x - 6)
6x2 + 58x + 80 = 02(40 + 29x + 3x2) = 02((8 + x)(5 + 3x)) = 0, ignore 28 + x = 0 => x = -85 + 3x = 0 => 3x = 5 => x= 1.666666....x = {-8, 1.66....}
It's a third-degree polynomial in 'x'. It's value depends on the value of 'x'. Every time 'x' changes, the value of the polynomial changes.
.58x = 63.4 x = 63.4/.58 x = 109.31
My best guess is that you have an exercise which is headed (and thus gives the context): "solve for x in the following equations", though it could quite easily be: solve for x in the following inequalities".As such, there is some sign between the 26 and 8x which unfortunately has been stripped out; it is one of {=≤≥}If I solve it for equals (=), then the others will follow:-58x - 26 = 8x - 230.6→ -26 + 230.6 = 58x + 8x→ 204.6 = 64x→ 3.196875 = xI have deliberately left the x on the right so that when the inequalities are inserted, the correct answer is given:-58x - 26 = 8x - 230.6 → 3.196875 = x, or with x on the left: x = 3.196875-58x - 26 < 8x - 230.6 → 3.196875 < x, or with x on the left: x > 3.196875-58x - 26 > 8x - 230.6 → 3.196875 > x, or with x on the left: x < 3.196875-58x - 26 ≤ 8x - 230.6 → 3.196875 ≤ x, or with x on the left: x ≥ 3.196875-58x - 26 ≥ 8x - 230.6 → 3.196875 ≥ x, or with x on the left: x ≤ 3.196875Note the change of the inequalities when the x is shifted from the right to the left so that the same meaning is read.
If d is the highest common factor of 58 and 145 then d equals 29:- So: 29 = 58x+145y Divide all terms by 29: 1 = 2x+5y Or: 2*1/4+5*1/10 = 1 Therefore the possible values of x and y are 1/4 and 1/10 respectively Check: 58*1/4 plus 145*1/10 = 29
is/of=%/10054/58=%/10058X%=5400x%=5400/58x%=93.103448 or about 93 percent
Algebraically. X = integers.X + (X + 1) = 592X + 1 = 592X = 58X = 29X + 1 = 30(29, 30)=======
x = 4/58x + 4 = 3x + 8Subtract 3x from both sides5x + 4 = 8Subtract 4 from both sides5x = 4Divide both sides by 5x = 4/5Check Work:8( 4/5 ) + 4 = 3( 4/5 ) + 88/1 ∙ 4/5 + 4 = 3/1 ∙ 4/5 + 832/5 + 20/5 = 12/5 + 40/5Remove the common denominator32 + 20 = 12 + 4052 = 52