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?

A nonlinear system has two equations sharing $x$ and $y$ , with a curve such as a square-root graph in at least one. Graph the original equations in Desmos and click their intersection, then compare its decimal coordinates with the exact answer choices. Watch the sign of the square root: squaring an equation can produce a candidate that fails the original.

Hints

Hint 1
The square root symbol gives a nonnegative value. Since it equals $y-1$ , what must be true of $y$ at any point that satisfies the system?
Hint 2
An intersection lies on both graphs, so its coordinates satisfy both equations. Type each equation as written, click where the graphs meet, and read the point in $x,y$ order.
Hint 3
Desmos labels the intersection with rounded decimals, while the choices use exact square roots. Evaluate both coordinates of a promising choice and compare them with the plotted point.

Step-by-step

Approach 1: Graph the original system
Step 1
Keep the square root's sign
The principal square root $\sqrt{3x-4}$ is never negative. It equals $y-1$ , so:
$y-1\ge0$
Add $1$ :
$y\ge1$
Squaring can erase this sign requirement.
Step 2
Find the point on both graphs
Type $x+2y=7$ and $\sqrt{3x-4}=y-1$ on separate Desmos lines. Click their intersection, the point on both graphs. Desmos labels it about $(2.056,2.472)$ . Those are rounded coordinates, not an exact radical answer.
Step 3
Match both exact coordinates
The plotted $x$ is a little above $2$ , and its $y$ is between $2$ and $3$ . Test the radical pair with positive $y$ : type $11-4\sqrt5$ and $-2+2\sqrt5$ on new lines. Desmos prints about $2.056$ and $2.472$ , matching both coordinates. So the ordered pair is $(11-4\sqrt5,-2+2\sqrt5)$ . Choice C.

Approach 2: Substitute and check the root's sign
Step 1
Write y in terms of x
The line lets you get $y$ alone for substitution, replacing it in the other equation. Subtract $x$ :
$2y=7-x$
Divide by $2$ :
$y=\frac{7-x}{2}$
Step 2
Make an equation in x
Replace $y$ in the square-root equation with $\frac{7-x}{2}$ :
$\sqrt{3x-4}=\frac{7-x}{2}-1$
Combine the terms on the right:
$\sqrt{3x-4}=\frac{5-x}{2}$
Step 3
Set the sign limit before squaring
A square root cannot be negative, so its equal on the right must obey:
$\frac{5-x}{2}\ge0$
Multiply by $2$ :
$5-x\ge0$
Subtract $5$ :
$-x\ge-5$
Multiply by $-1$ , flipping the inequality:
$x\le5$
Any later root above $5$ fails the original equation.
Step 4
Clear the square root
Square both whole sides to remove the radical:
$3x-4=\frac{(5-x)^2}{4}$
Multiply by $4$ :
$12x-16=(5-x)^2$
Expand the square:
$12x-16=x^2-10x+25$
Move every term to the right:
$0=x^2-22x+41$
Squaring may have added a false root, so keep the sign limit.
Step 5
Solve the squared equation exactly
Use completing the square: half of $-22$ is $-11$ , and $(-11)^2=121$ . Subtract $41$ :
$x^2-22x=-41$
Add $121$ to both sides:
$x^2-22x+121=80$
Rewrite the left side:
$(x-11)^2=80$
Take both square-root signs:
$x-11=\pm\sqrt{80}$
Since $80=16\cdot5$ , simplify:
$x-11=\pm4\sqrt5$
Add $11$ :
$x=11\pm4\sqrt5$
Step 6
Reject the extra root and find y
The plus-sign root $11+4\sqrt5$ exceeds $5$ , so it cannot satisfy the original square-root equation. Keep $x=11-4\sqrt5$ and substitute it into $y=\frac{7-x}{2}$ :
$y=\frac{7-(11-4\sqrt5)}{2}$
Simplify the numerator:
$y=\frac{-4+4\sqrt5}{2}$
Divide by $2$ :
$y=-2+2\sqrt5$
The pair is $(11-4\sqrt5,-2+2\sqrt5)$ . Choice C.

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

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

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