Question

$x \geq 0,$ \\(\

$$\begin{cases} 2^x + y = 3! + 10 \\\\ \\sqrt{y + 1} = x \\\\ x + y = ? \\end{cases}$$

\\)

Explanation:

Step1: Calculate 3!

$3! = 3×2×1 = 6$

Step2: Simplify first equation

$2^x + y = 6 + 10 = 16$

Step3: Express y from second equation

Square both sides: $y + 1 = x^2 \Rightarrow y = x^2 - 1$

Step4: Substitute y into first equation

$2^x + x^2 - 1 = 16 \Rightarrow 2^x + x^2 = 17$

Step5: Find x (x≥0)

Test x=4: $2^4 + 4^2 = 16 + 16 = 32≠17$; x=3: $8 + 9 =17$ → x=3

Step6: Calculate y

$y = 3^2 -1 =8$

Step7: Compute x+y

$3 +8=11$? Wait correction: Wait 2^3=8, 8+y=16 → y=8, yes x+y=3+8=11? Wait no wait step 4: 2^x + y=16, y=x²-1 so 2^x +x²-1=16 →2^x+x²=17, x=4:16+16=32 no;x=3:8+9=17 yes! So x=3,y=8,x+y=11. Yes correct.

Step1: Compute factorial value

$3! = 6$

Step2: Simplify first equation

$2^x + y = 6 + 10 = 16$

Step3: Express y via x

From $\sqrt{y+1}=x$, $y = x^2 -1$

Step4: Substitute y into equation

$2^x + x^2 -1 =16 \Rightarrow 2^x +x^2=17$

Step5: Solve for x (x≥0)

Test x=3: $2^3 +3^2=8+9=17$, so x=3

Step6: Find y value

$y=3^2 -1=8$

Step7: Calculate x+y

$3+8=11$

Answer:

7