Question

y=(x + 26)^2
y=x + 26
a solution to the given system of equations is (x,y). what is one possible value of x+26?
a -51
b -27
c 1
d 52

Explanation:

Step1: Set the two equations equal

Since \(y=(x + 26)^2\) and \(y=x + 26\), we set \((x + 26)^2=x + 26\). Let \(t=x + 26\), then the equation becomes \(t^2=t\).

Step2: Rearrange the equation

Move all terms to one - side: \(t^2-t=0\). Factor out \(t\): \(t(t - 1)=0\).

Step3: Solve for \(t\)

Using the zero - product property, if \(t(t - 1)=0\), then \(t = 0\) or \(t=1\).

Step4: Substitute back \(t=x + 26\)

When \(t = 0\), \(x+26=0\), so \(x=-26\). When \(t = 1\), \(x+26=1\), so \(x=-25\).

Step5: Find \(x + 26\) values

The values of \(x + 26\) are the values of \(t\), which are \(0\) and \(1\).

Answer:

C. 1