20 easy SAT Systems with nonlinear equations questions

These are the questions most test-takers get right. They are worth practising anyway: on the digital SAT the easy questions in Module 1 are what route you into the harder, higher-scoring Module 2, so dropping one costs more than it looks.

Every question below is a real item from the SAT Climb bank, tagged easy by the same difficulty model the app uses to build your practice. Pick an answer before you open the explanation.

Math · Advanced Math~2 per testEasy tier

None of these repeat the 25 hard Systems with nonlinear equations questions, so the two pages together are 45 different questions on this one skill.

Question 1Easy

$y = x - 4$ $y = x^{2} - 10$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $y_{1} + y_{2}$ ?

Show the answer and explanation

Why C is right

Substituting $y = x - 4$ into the second equation gives $x - 4$ = $x^{2}$ - 10, which simplifies to $x^{2}$ - $x - 6 = 0$ . Factoring yields $(x - 3) (x + 2) = 0$ , so $x_{1}$ = 3 and $x_{2}$ = -2. The corresponding y-values are $y_{1}$ = 3 - 4 = -1 and $y_{2}$ = -2 - 4 = -6. Therefore $y_{1} + y_{2}$ = -1 + (-6) = -7.

Why the others are wrong

AThis results from computing x₁ · x₂ = -6 instead of y₁ + y₂.
BThis is one of the intermediate values in the calculation, not the sum of both y-coordinates.
DThis results from incorrectly computing the sum with a sign error.

Question 2Easy

$y = -3x + 7$ $y = x^{2} + x - 5$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} + x_{2}$ ?

Show the answer and explanation

Why B is right

Substituting $y = -3x + 7$ into the second equation gives $-3x + 7$ = $x^{2}$ + $x - 5$ , which simplifies to $x^{2}$ + $4x - 12 = 0$ . By Vieta's formulas, the sum of the roots is the negative of the coefficient of x divided by the leading coefficient, so $x_{1} + x_{2}$ = -4.

Why the others are wrong

AThis results from incorrectly taking the positive value of the sum instead of the negative.
CThis is one of the x-coordinates from the factored form or a constant in the problem, not the sum.
DThis is the constant term from one of the equations, not the sum of the x-coordinates.

Question 3Easy

In the xy-plane, the graph of

y = (x - 1)^{2}

intersects the graph of

y = 3x + 1

at two points.

What is the solution (x, y) with the greater y-value to this system of equations?

Show the answer and explanation

Why B is right

Setting $(x - 1)^{2} = 3x + 1$ gives $x^{2}$ - $2x + 1 = 3x + 1$ , which simplifies to $x^{2}$ - $5x = 0$ . Factoring yields $x(x - 5) = 0$ , so $x = 0$ or $x = 5$ . When $x = 5$ , $y = 3$ (5) + 1 = 16. The solution with the greater y-value is (5, 16).

Why the others are wrong

AThis results from an arithmetic error in solving the quadratic, obtaining x = 4 instead of x = 5.
CThis includes a sign error when solving x(x - 5) = 0, obtaining x = -5 instead of x = 5.
DThis uses the correct x-value but drops a term when computing y, obtaining 14 instead of 16.

Question 4Easy

$y = 4x$ $x^{2} + y^{2} = 68$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $y_{1} + y_{2}$ ?

Show the answer and explanation

Why D is right

Substituting $y = 4x$ into the circle equation gives $x^{2}$ + $(4x)^{2} = 68$ , which simplifies to $x^{2}$ + $16x^{2} = 68$ or $17x^{2} = 68$ , so $x^{2}$ = 4 and $x = \pm 2$ . The corresponding y-values are $y_{1}$ = $4(2) = 8$ and $y_{2}$ = $4(-2) = -8$ . Therefore $y_{1} + y_{2}$ = $8 + (-8) = 0$ .

Why the others are wrong

AThis represents finding only one y-value instead of computing the sum.
BThis represents the coefficient from the linear equation instead of the sum of y-values.
CThis results from a sign error, possibly adding absolute values instead of the actual y-values.

Question 5Easy

$y = x^{2} - 3$ $y = x - 3$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $y_{1} + y_{2}$ ?

Show the answer and explanation

Why B is right

Substituting $y = x - 3$ into the first equation gives $x - 3$ = $x^{2}$ - 3, which simplifies to $x^{2}$ - $x = 0$ or $x(x - 1) = 0$ . The solutions are $x = 0$ and $x = 1$ . The corresponding y-values are $y_{1}$ = 0 - 3 = -3 and $y_{2}$ = 1 - 3 = -2. Therefore $y_{1} + y_{2}$ = -3 + (-2) = -5.

Why the others are wrong

AThis results from incorrectly doubling one of the y-values instead of summing both.
CThis is only one of the two y-coordinates, not their sum.
DThis results from an error in sign when computing the sum.

Question 6Easy

$x^{2} + y^{2} = 25$ $y = 4$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of | $x_{1} - x_{2}$ |?

Show the answer and explanation

Why D is right

Substituting $y = 4$ into $x^{2}$ + $y^{2}$ = 25 gives $x^{2}$ + 16 = 25, so $x^{2}$ = 9. The solutions are $x_{1}$ = 3 and $x_{2}$ = -3. Therefore | $x_{1} - x_{2}$ | = |3 - (-3)| = |6| = 6.

Why the others are wrong

AThis represents finding only one x-value (3) instead of computing the absolute difference between both solutions.
BThis incorrectly uses x² = 9 directly as the answer instead of finding the difference of the actual x-values.
CThis results from computing 3 - (-3) = 6 but then incorrectly applying a negative sign, forgetting absolute value makes the result positive.

Question 7Easy

In the xy-plane, the graph of

y = x^{2} - 6x + 5

intersects the graph of

y = -3

at two points.

What is the solution (x, y) with the smaller x-value to this system of equations?

Show the answer and explanation

Why A is right

Setting $x^{2}$ - $6x + 5 = -3$ gives $x^{2}$ - $6x + 8 = 0$ , which factors as $(x - 2) (x - 4) = 0$ . The solutions are $x = 2$ and $x = 4$ . The smaller x-value is $x = 2$ , giving the point (2, -3).

Why the others are wrong

BThis results from a sign error when solving the quadratic equation, obtaining x = -2 instead of x = 2.
CThis uses the correct x-value but drops the constant term when computing y, obtaining 5 instead of -3.
DThis results from an arithmetic error in factoring, obtaining x = 1 as a factor instead of x = 2.

Question 8Easy

$y = x - 2$ $x^{2} + y^{2} = 100$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} \cdot x_{2}$ ?

Show the answer and explanation

Why A is right

Substituting $y = x - 2$ into the circle equation gives $x^{2}$ + $(x - 2)^{2} = 100$ . Expanding yields $x^{2}$ + $x^{2}$ - $4x + 4 = 100$ , which simplifies to $2x^{2} - 4x - 96 = 0$ or $x^{2}$ - $2x - 48 = 0$ . By Vieta's formulas, the product of the roots is -48.

Why the others are wrong

BThis results from a sign error when applying Vieta's formulas or computing the product.
CThis represents finding only one x-value instead of computing the product of both.
DThis represents the coefficient from the linear term instead of the product of solutions.

Question 9Easy

$y = x - 1$ $y = x^{2} - 7x + 11$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $y_{1} + y_{2}$ ?

Show the answer and explanation

Why A is right

Substituting $y = x - 1$ into the second equation gives $x - 1$ = $x^{2}$ - $7x + 11$ , which simplifies to $x^{2}$ - $8x + 12 = 0$ . Using Vieta's formulas, $x_{1} + x_{2}$ = 8. Since $y = x - 1$ , we have $y_{1} + y_{2}$ = ( $x_{1}$ - 1) + ( $x_{2}$ - 1) = $x_{1} + x_{2}$ - 2 = 8 - 2 = 6.

Why the others are wrong

BThis incorrectly uses x₁ + x₂ instead of y₁ + y₂.
CThis represents only one y-coordinate value rather than the sum of both.
DThis results from a sign error in computing the sum.

Question 10Easy

$y = x + 8$ $y = x^{2} - 2x$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} \cdot x_{2}$ ?

Show the answer and explanation

Why A is right

Substituting $y = x + 8$ into the second equation gives $x + 8$ = $x^{2}$ - 2x, which simplifies to $x^{2}$ - $3x - 8 = 0$ . By Vieta's formulas, the product of the roots is the constant term divided by the leading coefficient, so $x_{1} \cdot x_{2}$ = -8.

Why the others are wrong

BThis results from incorrectly taking the positive value of the product instead of the negative.
CThis is the sum of the x-coordinates (x₁ + x₂ = 3 from Vieta's formulas), not their product.
DThis is one of the x-coordinates from solving the quadratic, not the product of both.

Question 11Easy

In the xy-plane, the graph of

y = x^{2} + 2x

intersects the graph of

y = 8

at two points.

What is the solution (x, y) with the greater x-value to this system of equations?

Show the answer and explanation

Why C is right

Setting $x^{2}$ + $2x = 8$ gives $x^{2}$ + $2x - 8 = 0$ , which factors as $(x + 4) (x - 2) = 0$ . The solutions are $x = -4$ and $x = 2$ . The greater x-value is $x = 2$ , giving the point (2, 8).

Why the others are wrong

AThis uses the smaller x-value solution x = -4 instead of the greater x-value.
BThis results from an arithmetic error in factoring, obtaining x = 3 as a factor instead of x = 2.
DThis uses the correct x-value but drops a term when computing y, obtaining 6 instead of 8.

Question 12Easy

$y = x + 6$ $y = x^{2} - 36$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} + x_{2}$ ?

Show the answer and explanation

Why D is right

Substituting $y = x + 6$ into the second equation gives $x + 6$ = $x^{2}$ - 36, which simplifies to $x^{2}$ - $x - 42 = 0$ . Factoring yields $(x - 7) (x + 6) = 0$ , so $x = 7$ or $x = -6$ . Therefore $x_{1} + x_{2}$ = 7 + (-6) = 1.

Why the others are wrong

AThis represents finding only one x-coordinate instead of computing the sum.
BThis results from a sign error when computing the sum.
CThis represents using only the negative solution or making an error in the sum calculation.

Question 13Easy

$y = 2x - 1$ $y = x^{2} - 4x + 3$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} \cdot x_{2}$ ?

Show the answer and explanation

Why D is right

Setting the expressions equal gives $2x - 1$ = $x^{2}$ - $4x + 3$ . Rearranging yields $x^{2}$ - $6x + 4 = 0$ . By Vieta's formulas, the product of roots is $c/a = 4/1 = 4$ . Alternatively, solving gives $\displaystyle x = 3 \pm \sqrt{5}$ and $\displaystyle (3 + \sqrt{5}) (3 - \sqrt{5}) = 9 - 5 = 4$ .

Why the others are wrong

AThis results from confusing the product with the sum of y-coordinates or another calculation error.
BThis results from a sign error when applying Vieta's formulas or rearranging the equation.
CThis results from finding only one solution and using that value as the product.

Question 14Easy

$y = 5$ $y = x^{2} + 2x - 20$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} \cdot x_{2}$ ?

Show the answer and explanation

Why A is right

Substituting $y = 5$ into the second equation gives 5 = $x^{2}$ + $2x - 20$ , which simplifies to $x^{2}$ + $2x - 25 = 0$ . By Vieta's formulas, the product of the roots is the constant term divided by the leading coefficient, so $x_{1} \cdot x_{2}$ = -25.

Why the others are wrong

BThis results from incorrectly taking the absolute value of the product or missing the negative sign.
CThis is the y-value of both solutions, not the product of the x-coordinates.
DThis is the sum of the x-coordinates (x₁ + x₂ = -2 from Vieta's formulas), not their product.

Question 15Easy

In the xy-plane, the graph of y = -

2x^{2} + 10

intersects the graph of

y = 2

at two points.

What is the solution (x, y) with the positive x-value to this system of equations?

Show the answer and explanation

Why D is right

Setting $-2x^{2} + 10 = 2$ gives $-2x^{2} = -8$ , so $x^{2}$ = 4. Taking the square root yields $x = \pm 2$ . The positive x-value is $x = 2$ , giving the point $(2, 2)$ .

Why the others are wrong

AThis results from an arithmetic error when solving x² = 4, obtaining x = 3 instead of x = 2.
BThis uses the negative x-value solution x = -2 instead of the positive x-value.
CThis uses the correct x-value but incorrectly uses the y-intercept of the parabola instead of y = 2.

Question 16Easy

$y = 3x + 2$ $y = x^{2} - 6$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} \cdot x_{2}$ ?

Show the answer and explanation

Why A is right

Substituting $y = 3x + 2$ into the second equation gives $3x + 2$ = $x^{2}$ - 6, which simplifies to $x^{2}$ - $3x - 8 = 0$ . By Vieta's formulas, the product of the roots is the constant term divided by the leading coefficient, which is -8/1 = -8.

Why the others are wrong

BThis results from computing the sum of the roots instead of the product.
CThis results from a sign error when applying Vieta's formulas or rearranging the equation.
DThis is the x-coordinate of only one solution rather than the product of both.

Question 17Easy

$y = x^{2} - 4$ $y = 2x + 5$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} + x_{2}$ ?

Show the answer and explanation

Why B is right

Substituting $y = 2x + 5$ into the first equation yields $2x + 5$ = $x^{2}$ - 4. Rearranging gives $x^{2}$ - $2x - 9 = 0$ . By Vieta's formulas, the sum of the roots is the opposite of the coefficient of x divided by the leading coefficient, which is -(-2)/1 = 2.

Why the others are wrong

AThis results from incorrectly applying Vieta's formulas with the wrong sign, treating the sum as -2 instead of 2.
CThis is the x-coordinate of only one solution, not the sum of both x-coordinates.
DThis is the sum of the y-coordinates y₁ + y₂, not the sum of the x-coordinates.

Question 18Easy

$y = x + 5$ $y = x^{2} - 3$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} + x_{2}$ ?

Show the answer and explanation

Why B is right

Substituting the first equation into the second gives $x + 5$ = $x^{2}$ - 3, which simplifies to $x^{2}$ - $x - 8 = 0$ . By Vieta's formulas, the sum of the roots is the negative of the coefficient of x divided by the coefficient of $x^{2}$ , which is -(-1)/1 = 1.

Why the others are wrong

AThis results from incorrectly taking the sum as -1/1 instead of -(-1)/1.
CThis is the x-coordinate of only one solution, not the sum of both x-coordinates.
DThis is the sum of the y-coordinates, y₁ + y₂, not the sum of the x-coordinates.

Question 19Easy

$y = 4x + 1$ $y = x^{2} + 5x - 2$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} + x_{2}$ ?

Show the answer and explanation

Why B is right

Substituting $y = 4x + 1$ into the second equation gives $4x + 1$ = $x^{2}$ + $5x - 2$ , which simplifies to $x^{2}$ + $x - 3 = 0$ . By Vieta's formulas, the sum of the roots is -1/1 = -1.

Why the others are wrong

AThis results from a sign error when applying Vieta's formulas or when rearranging the equation.
CThis is the x-coordinate of only one solution rather than the sum of both.
DThis results from adding the y-coordinates instead of the x-coordinates or from reading the coefficient incorrectly.

Question 20Easy

$y = x^{2} + 3x - 10$ $y = x + 2$ The system of equations above has solutions $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ . What is the value of $x_{1} x_{2}$ ?

Show the answer and explanation

Why A is right

Substituting $y = x + 2$ into the first equation yields $x + 2$ = $x^{2}$ + $3x - 10$ . Rearranging gives $x^{2}$ + $2x - 12 = 0$ . By Vieta's formulas, the product of the roots is the constant term divided by the leading coefficient, which is -12/1 = -12.

Why the others are wrong

BThis is the constant term from the original quadratic equation, not the product of the roots after substitution.
CThis is the sum of the x-coordinates x₁ + x₂, not their product.
DThis results from incorrectly applying Vieta's formulas with the wrong sign.

What to do after easy

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Questions are written by SAT Climb and drawn from its own item bank. SAT® is a registered trademark of College Board, which is not affiliated with and does not endorse SAT Climb.

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