There is no unique solution — this equation is true for all values of ddd.
Explanation
The given equation is: 5d+2(2−d)=3(1+d)+15d + 2(2 – d) = 3(1 + d) + 15d+2(2−d)=3(1+d)+1
To solve for ddd, we start by applying the distributive property, simplifying each side of the equation.
On the left-hand side, 2(2−d)2(2 – d)2(2−d) becomes 4−2d4 – 2d4−2d. Adding this to 5d5d5d gives: 5d+4−2d=3d+45d + 4 – 2d = 3d + 45d+4−2d=3d+4
On the right-hand side, 3(1+d)3(1 + d)3(1+d) becomes 3+3d3 + 3d3+3d, and adding 1 gives: 3+3d+1=3d+43 + 3d + 1 = 3d + 43+3d+1=3d+4
After simplifying both sides, we get the same expression: 3d+4=3d+43d + 4 = 3d + 43d+4=3d+4
This equation tells us that both sides are identical. When we try to isolate the variable ddd, we subtract 3d+43d + 43d+4 from both sides: (3d+4)−(3d+4)=0⇒0=0(3d + 4) – (3d + 4) = 0 \Rightarrow 0 = 0(3d+4)−(3d+4)=0⇒0=0
This result is a true statement and does not depend on the value of ddd. It implies that the original equation is always true, regardless of the value of ddd. Therefore, every real number is a solution. Such equations are called identities.
In contrast, if the result had been a false statement (like 0=50 = 50=5), it would mean there is no solution. If we had isolated ddd and found a specific value (like d=2d = 2d=2), the equation would have had one solution.
Here, since the equation is always true, the solution is:
