Which ordered pair satisfies the system of equations above?

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SAT Nonlinear Equations & Systems Question 93 | Free SAT Prep | Aniko

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Question 93·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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\begin{cases} x + 2y = 7 \\ \sqrt{3x - 4} = y - 1 \end{cases}

Which ordered pair $(x,\,y)$ satisfies the system of equations above?

For systems that mix a linear equation with a square-root equation, use substitution to express one variable from the linear equation and plug it into the square-root equation so you have a single equation in one variable. Isolate the square root, apply the fact that square roots are nonnegative to restrict possible solutions, then square both sides and simplify to a quadratic. After solving the quadratic, always check each candidate solution in the original (unsquared) equation to discard any extraneous solutions before computing the corresponding second coordinate.

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Start with substitution

Use the linear equation $x + 2y = 7$ to solve for $y$ in terms of $x$ . Then substitute this expression for $y$ into the other equation.

Focus on the square root equation

After substitution, you will get an equation of the form $\sqrt{3x - 4} = \text{(expression in } x\text{)}$ . Remember a square root is always nonnegative; what does that tell you about the expression on the right?

Remove the square root carefully

Isolate the square root, then square both sides to get rid of it. Simplify to a quadratic, solve for $x$ , and then check which $x$ actually satisfies the original (unsquared) equation.

Don’t forget to find y

Once you have the correct $x$ -value, plug it back into the simpler linear equation $x + 2y = 7$ to find the corresponding $y$ -value.

Desmos Guide

Graph the linear equation

Enter the first equation in Desmos as y = (7 - x)/2. This is the line representing all solutions to $x + 2y = 7$ .

Graph the square-root equation as a function

Rewrite the second equation $\sqrt{3x - 4} = y - 1$ as $y = 1 + \sqrt{3x - 4}$ and enter it in Desmos as y = 1 + sqrt(3x - 4). This graphs the curve defined by the square-root equation.

Find the intersection point

Look for the point where the line y = (7 - x)/2 and the curve y = 1 + sqrt(3x - 4) intersect. Tap that intersection; the coordinates shown by Desmos are the $(x, y)$ pair that solves the system.

Step-by-step Explanation

Express y in terms of x

From the first equation,

x + 2y = 7,

solve for $y$ :

2y = 7 - x \\[0.7em] y = \frac{7 - x}{2}.

Use this in the second equation $\sqrt{3x - 4} = y - 1$ :

\sqrt{3x - 4} = \frac{7 - x}{2} - 1 = \frac{5 - x}{2}.

Now you have an equation with only $x$ .

Use the square root properties and set up a quadratic

Because a square root is always nonnegative, the right side $\frac{5 - x}{2}$ must be greater than or equal to $0$ , so

5 - x \ge 0 \quad \Rightarrow \quad x \le 5.

Now square both sides of

\sqrt{3x - 4} = \frac{5 - x}{2}

to remove the square root:

3x - 4 = \left(\frac{5 - x}{2}\right)^2.

Multiply both sides by $4$ to clear the denominator:

4(3x - 4) = (5 - x)^2 \\[0.7em] 12x - 16 = (5 - x)^2.

Simplify to a quadratic and solve for x

Expand the right side:

(5 - x)^2 = x^2 - 10x + 25.

Set this equal to $12x - 16$ :

x^2 - 10x + 25 = 12x - 16.

Bring all terms to one side:

x^2 - 10x + 25 - 12x + 16 = 0 \\[0.7em] x^2 - 22x + 41 = 0.

Use the quadratic formula with $a = 1$ , $b = -22$ , $c = 41$ :

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{22 \pm \sqrt{22^2 - 4\cdot1\cdot41}}{2}.

Compute the discriminant:

22^2 - 4\cdot41 = 484 - 164 = 320 = 64\cdot5,

so $\sqrt{320} = 8\sqrt{5}$ , and

x = \frac{22 \pm 8\sqrt{5}}{2} = 11 \pm 4\sqrt{5}.

Choose the valid x and find y

From the square-root equation we had $x \le 5$ . Since $\sqrt{5}$ is a bit more than $2$ , $4\sqrt{5}$ is a bit more than $8$ , so

$11 + 4\sqrt{5} > 11 + 8 = 19 > 5$ (this violates $x \le 5$ ),
$11 - 4\sqrt{5}$ is between $11 - 9 = 2$ and $11 - 8 = 3$ , so it is less than $5$ and allowed.

Thus the valid solution for $x$ is $x = 11 - 4\sqrt{5}$ . Now use $y = \frac{7 - x}{2}$ :

y = \frac{7 - (11 - 4\sqrt{5})}{2} = \frac{-4 + 4\sqrt{5}}{2} = -2 + 2\sqrt{5}.

So the ordered pair that satisfies both equations is $(11-4\sqrt{5},\,-2+2\sqrt{5})$ .

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Step-by-step Explanation

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