Question 150·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Consider the system of equations

\begin{cases} y = (x - 3)^2 + 1 \\ x + y = 7 \end{cases}

Which of the following is a solution for $x$ in the system above?

A nonlinear system has two equations in the same variables and at least one curved graph. A squared expression signals that the curve may meet a line twice. Graph both equations in Desmos and click every intersection. Read the first coordinate for $x$ , then compare it with the exact choices as decimals. Don’t use the $y$ -coordinate or stop at the first crossing.

Hints

Hint 1
A solution to a system must work in both equations. Graph each equation as written; the shared points show where both conditions hold. Could the line cross the curve more than once?
Hint 2
An intersection is labeled $(x,y)$ , with $x$ first and $y$ second. A parabola is the U-shaped graph of a squared expression. Check both crossings before reading their $x$ -coordinates.
Hint 3
An exact radical contains a square root, but Desmos labels clicked points with rounded decimals. Evaluate a candidate expression in Desmos and compare it with an intersection’s $x$ -coordinate.

Step-by-step

Approach 1: Graph the system and match a choice
Step 1
Find both shared points
Type $y=(x-3)^2+1$ and $x+y=7$ on separate Desmos lines. Click both intersections. Desmos labels them approximately $(0.697,6.303)$ and $(4.303,2.697)$ . Each intersection is a solution to the system because it lies on both graphs. The first coordinate is $x$ , so the possible $x$ -values are about $0.697$ and $4.303$ .
Step 2
Match the exact expression
A radical is an expression containing a square root. Type $\frac{5+\sqrt{13}}{2}$ on a new Desmos line. It prints about $4.303$ , matching the first coordinate of the right-hand intersection. Dividing that numerator by $4$ instead would halve the value and miss both intersections. So one possible $x$ -value is $\frac{5+\sqrt{13}}{2}$ . Choice C.

Approach 2: Find the exact values by substitution
Step 1
Get $y$ by itself
Subtract $x$ from both sides of $x+y=7$ to isolate $y$ , meaning leave it alone:
$y=7-x.$
Step 2
Make one equation in $x$
At a shared point, both equations give the same $y$ . Substitution means replacing a variable with an equal expression. Replace $y$ in the curved equation with $7-x$ :
$7-x=(x-3)^2+1.$
Step 3
Expand the square
Multiply $(x-3)(x-3)$ . Its two cross-products are $-3x$ each, so the middle term is $-6x$ . Expand the right side:
$7-x=x^2-6x+9+1.$
Step 4
Put all terms on one side
A quadratic is an equation containing $x^2$ . To prepare this one for the solving formula, make one side zero. Combine $9+1$ : $7-x=x^2-6x+10$ . Add $x$ to both sides: $7=x^2-5x+10$ . Subtract $7$ from both sides:
$0=x^2-5x+3.$
Step 5
Apply the quadratic formula
For $ax^2+bx+c=0$ , the quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . Here $a=1$ , $b=-5$ , and $c=3$ . Keep the negative sign on $b$ when you substitute:
$x=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(3)}}{2(1)}.$
Step 6
Simplify and choose a possible value
Evaluate the numbers in the formula: $x=\frac{5\pm\sqrt{25-12}}{2}$ . Subtract under the square root: $x=\frac{5\pm\sqrt{13}}{2}$ . The $\pm$ gives two possible $x$ -values. The plus-sign value is $\frac{5+\sqrt{13}}{2}$ , which appears among the choices. Choice C.

Question 150·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

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