Integers do not have sides so the question does not make sense.
Let the first of the three odd consecutive integers be x, so that the second of these integers would be x + 2, and the third one would be x + 4. We have: 3x = 2[(x + 2)+ (x + 4)] + 3 3x = 2(2x + 6) +3 3x = 4x + 12 + 3 (subtract 4x to both sides) -x = 15 (multiply by -1 to both sides) x = -15 (the first one) So the integers are -15, -13, and -11. The average of those integers is (-15 + -13 + -11)/2 = -39/2 = -19.5.
2x + 22 = x Subtract x from both sides: x + 22 = 0 Subtract 22 from both sides: x = -22
There are 18, but only 7 primitive ones. The rest are similar to the primitive triangles : for example (15, 50, 25) is similar to (3, 4, 5).
First you subtract 3x from both sides and get -7=2x+3. Then subtract 3 from both sides and get -10=2x. Then divide both sides by 2 and x=-5.
70 = 81-n subtract 70 from both sides 0 = 11 - n subtract 11 from both sides -11 = -n divide both sides by -1 11 = n
Let x be the smallest of the consecutive odd integers. Since consecutive odd integers differ by 2, we havex + (x + 2) + (x + 4) + (x + 6) = -2204x + 12 = -220 (subtract 12 to both sides)4x = -232 (divide by 4 to both sides)x = -58Thus, the four consecutive odd integers whose sum is -220 are -58, -56, -54, and -52.
Algebraically, X = integers. X + (X + 1) = 237 gather all terms on the left 2X + 1 = 237 subtract 1 from each side 2X = 236 divide both sides integers by 2 X = 118 --------------so, X + 1 = 119 ----------------so, The two consecutive integers that = 237 are 118 and 119 -------------------
That is the correct spelling of "scalene" (in one sense a triangle with three dissimilar sides).
Yes. When you subtract two, you get two congruent sides.
Let the odd integers be , n, n+2 & n+4 Hence n + (n+2) + (n +4) = 471 Add LHS 3n + 6 = 471 Subtract '6' from both sides 3n = 465 Divide both sides by '3' n = 155 n + 2 = 157 n + 4 = 159 Hence the integers are 155,157 & 159.
9Y + 4 = 2Y + 25 subtract 2Y from both sides 7Y + 4 = 25 subtract 4 from both sides 7Y = 21 divide both sides integers by 7 (7/7)Y = 21/7 Y = 3 -----------------check in original equation 9(3) + 4 = 2(3) + 25 27 + 4 = 6 + 25 31 = 31 ---------------checks
Let the first of the three odd consecutive integers be x, so that the second of these integers would be x + 2, and the third one would be x + 4. We have: 3x = 2[(x + 2)+ (x + 4)] + 3 3x = 2(2x + 6) +3 3x = 4x + 12 + 3 (subtract 4x to both sides) -x = 15 (multiply by -1 to both sides) x = -15 (the first one) So the integers are -15, -13, and -11. The average of those integers is (-15 + -13 + -11)/2 = -39/2 = -19.5.
2x + 22 = x Subtract x from both sides: x + 22 = 0 Subtract 22 from both sides: x = -22
There are 18, but only 7 primitive ones. The rest are similar to the primitive triangles : for example (15, 50, 25) is similar to (3, 4, 5).
First you subtract 3x from both sides and get -7=2x+3. Then subtract 3 from both sides and get -10=2x. Then divide both sides by 2 and x=-5.
70 = 81-n subtract 70 from both sides 0 = 11 - n subtract 11 from both sides -11 = -n divide both sides by -1 11 = n
You can add or subtract any quantity on both sides of an equation, without changing the equation's solution set. Just make sure you add or subtract the same thing on both sides.