We solve the equation, ( a + a + 1 )^2 = 112 + a^2 + ( a + 1 )^2;
Then, ( 2a + 1 )^2 = 112 + a^2 + a^2 + 2a +1;
4a^2 + 4a + 1 = 113 + 2a^2 + 2a;
Finally, 2a^2 + 2a - 112 = 0;
a^2 + a - 56 = 0;
We use <span>Quadratic Formula for this Quadratic Equation;
The solutions are a1 = 7 and a2 = -8;
But a is a natural number; so, a = 7;
The natural consecutive numbers are 7 and 8.</span>
Check your other post. I answered there!
Answer:
The solution is: (x, y) = (9, -7)
Step-by-step explanation:
Step 1: Solve one of the equations for either x = or y = . We will solve second equation for y.
x+y=3
y=3-x
Step 2: Substitute the solution from step 1 into the second equation.
3x+4(3-x)=5
Step 3: Solve this new equation.
3x+4(3-x)=5
3x+12-4x=3
-x+12=3
-x=3-12
x=9
Step 4: Solve for the second variable
y=3-x
y=3-9
y=-7
The solution is: (x, y) = (9, -7)
