20 medium SAT Systems with nonlinear equations questions

Medium is where most scores are actually won and lost. These questions are not tricky for the sake of it, but every one of them has a wrong answer built to catch a specific shortcut.

Every question below is a real item from the SAT Climb bank, tagged medium 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 testMedium 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 1Medium

y=x2−6x+5y = x^{2} - 6x + 5 y=2x−7y = 2x - 7 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1+x2x_{1} + x_{2}?

Show the answer and explanation

Why B is right

Substituting y=2x−7y = 2x - 7 into the first equation gives 2x−72x - 7 = x2x^{2} - 6x+56x + 5. Rearranging yields x2x^{2} - 8x+12=08x + 12 = 0. By Vieta's formulas, the sum of the roots equals the negative of the coefficient of x divided by the leading coefficient, which is 8.

Why the others are wrong

  • AThis results from incorrectly taking the coefficient -6 from the original quadratic instead of the rearranged form.
  • CThis is the value of x₁ only, not the sum of both x-coordinates.
  • DThis results from incorrectly adding y₁ + y₂ instead of x₁ + x₂.
Question 2Medium

x2+y2=52x^{2} + y^{2} = 52 y=x−4y = x - 4 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

Show the answer and explanation

Why C is right

Substituting y=x−4y = x - 4 into the first equation yields x2x^{2} + (x−4)2=52(x - 4)^{2} = 52. Expanding gives x2x^{2} + x2x^{2} - 8x+16=528x + 16 = 52, so 2x2−8x−36=02x^{2} - 8x - 36 = 0, or x2x^{2} - 4x−18=04x - 18 = 0. By Vieta's formulas, x1⋅x2x_{1} \cdot x_{2} = -18. Since y=x−4y = x - 4, we have y1⋅y2y_{1} \cdot y_{2} = ([MATH]x1−4)([MATH]x_{1} - 4)(x_{2}[/MATH] - 4) = x1x2x_{1} x_{2} - 4(x1+x2)4(x_{1} + x_{2}) + 16 = -18 - 4(4) + 16 = -18 - 16 + 16 = -18.

Why the others are wrong

  • AThis results from finding the sum of the y-coordinates instead of their product.
  • BThis results from an incorrect sign when computing the product.
  • DThis is only one of the y-values, not the product of both.
Question 3Medium

y=−x2+8x−12y = - x^{2} + 8x - 12 y=x+2y = x + 2 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

Show the answer and explanation

Why B is right

Substituting y=x+2y = x + 2 into the first equation gives x+2x + 2 = -x2x^{2} + 8x−128x - 12. Rearranging yields x2x^{2} - 7x+14=07x + 14 = 0. By Vieta's formulas, x1⋅x2x_{1} \cdot x_{2} = 14. Since y=x+2y = x + 2, we have y1⋅y2y_{1} \cdot y_{2} = ([MATH]x1+2)([MATH]x_{1} + 2)(x_{2}[/MATH] + 2) = x1x2x_{1} x_{2} + 2(x1+x2)2(x_{1} + x_{2}) + 4 = 14 + 2(7) + 4 = 30.

Why the others are wrong

  • AThis is the value of x₁ · x₂ rather than y₁ · y₂.
  • CThis results from a sign error in the calculation of y₁ · y₂.
  • DThis is the value of one y-coordinate rather than the product of both y-coordinates.
Question 4Medium

y=x2+4x−5y = x^{2} + 4x - 5 y=3x+7y = 3x + 7 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1⋅x2x_{1} \cdot x_{2}?

Show the answer and explanation

Why A is right

Substituting y=3x+7y = 3x + 7 into the first equation gives 3x+73x + 7 = x2x^{2} + 4x−54x - 5. Rearranging yields x2x^{2} + x−12=0x - 12 = 0. By Vieta's formulas, the product of the roots equals the constant term divided by the leading coefficient, which is -12/1 = -12.

Why the others are wrong

  • BThis results from an incorrect sign when applying Vieta's formulas to the constant term.
  • CThis results from finding only one solution instead of computing the product of both solutions.
  • DThis is the product of the y-coordinates, y₁ · y₂, not the x-coordinates.
Question 5Medium

y=x2−10x+21y = x^{2} - 10x + 21 y=−x+9y = -x + 9 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

Show the answer and explanation

Why A is right

Substituting y=−x+9y = -x + 9 into the first equation yields −x+9-x + 9 = x2x^{2} - 10x+2110x + 21. Rearranging gives x2x^{2} - 9x+12=09x + 12 = 0. By Vieta's formulas, x1⋅x2x_{1} \cdot x_{2} = 12. Since y=−x+9y = -x + 9, we have y1⋅y2y_{1} \cdot y_{2} = (-x1x_{1} + 9)(-x2x_{2} + 9) = x1x2x_{1} x_{2} - 9(x1+x2)9(x_{1} + x_{2}) + 81 = 12 - 9(9) + 81 = 12 - 81 + 81 = 12.

Why the others are wrong

  • BThis is the product of the x-coordinates x₁ · x₂ computed incorrectly.
  • CThis results from a sign error when expanding the product y₁ · y₂.
  • DThis results from finding only one solution y = 6 and not finding the second solution.
Question 6Medium

y=x2−8x+7y = x^{2} - 8x + 7 y=−x+1y = -x + 1 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of (x1−x2)2(x₁ - x₂)^{2}?

Show the answer and explanation

Why D is right

Substituting y=−x+1y = -x + 1 into the first equation yields −x+1-x + 1 = x2x^{2} - 8x+78x + 7, which simplifies to x2x^{2} - 7x+6=07x + 6 = 0. The sum of roots is 7 and product is 6. Using (x1−x2)2(x_{1} - x_{2}) ^{2} = (x1+x2)2(x_{1} + x_{2}) ^{2} - 4x1x2x_{1} x_{2}, we get (7)2−4(7)^{2} - 4(6) = 49 - 24 = 25.

Why the others are wrong

  • AThis is the sum of the x-coordinates, not (x₁ - x₂)².
  • BThis is the negative of the correct value, resulting from a sign error in calculation.
  • CThis results from finding only one solution or incorrectly computing the difference as (x₁ - x₂) = 1.
Question 7Medium

y=x2+2x−8y = x^{2} + 2x - 8 y=x+4y = x + 4 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1⋅x2x_{1} \cdot x_{2}?

Show the answer and explanation

Why B is right

Substituting y=x+4y = x + 4 into the first equation yields x+4x + 4 = x2x^{2} + 2x−82x - 8. Rearranging gives x2x^{2} + x−12=0x - 12 = 0. By Vieta's formulas, the product of the roots is -12/1 = -12.

Why the others are wrong

  • AThis is the product of the y-coordinates minus a constant, not the product of x-coordinates.
  • CThis results from a sign error when applying Vieta's formulas to the constant term.
  • DThis results from finding only one solution instead of computing the product of both solutions.
Question 8Medium

y=x2−10x+21y = x^{2} - 10x + 21 y=−2x+9y = -2x + 9 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x12+x22x_{1} ^{2} + x _{2} ^{2}?

Show the answer and explanation

Why A is right

Substituting y=−2x+9y = -2x + 9 into the first equation yields −2x+9-2x + 9 = x2x^{2} - 10x+2110x + 21, which simplifies to x2x^{2} - 8x+12=08x + 12 = 0. By Vieta's formulas, x1+x2x_{1} + x_{2} = 8 and x1⋅x2x_{1} \cdot x_{2} = 12. Using the identity x12x_{1} ^{2} + x22x_{2} ^{2} = (x1+x2)2(x_{1} + x_{2}) ^{2} - 2x1x2x_{1} x_{2}, we get 82−28^{2} - 2(12) = 64 - 24 = 40.

Why the others are wrong

  • BThis results from incorrectly computing (x₁ + x₂)² without subtracting 2x₁x₂.
  • CThis is the value of (x₁ + x₂)² only, without applying the full identity.
  • DThis is the value of one x² only, not the sum of both squares.
Question 9Medium

y=x2−10x+21y = x^{2} - 10x + 21 y=−3y = -3 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1⋅x2x_{1} \cdot x_{2}?

Show the answer and explanation

Why D is right

Substituting y=−3y = -3 into the first equation yields -3 = x2x^{2} - 10x+2110x + 21, which simplifies to x2x^{2} - 10x+24=010x + 24 = 0. By Vieta's formulas, the product of the roots equals the constant term divided by the coefficient of x2x^{2}, which is 24/1 = 24.

Why the others are wrong

  • AThis is the sum of the x-coordinates, not their product.
  • BThis results from finding only one solution (x = 4) and not recognizing there is a second solution.
  • CThis is the negative of the correct product, resulting from a sign error when applying Vieta's formulas.
Question 10Medium

x2+y2=49x^{2} + y^{2} = 49 x=7x = 7 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

Show the answer and explanation

Why B is right

Substituting x=7x = 7 into the first equation yields 49 + y2y^{2} = 49, which simplifies to y2y^{2} = 0. This gives y=0y = 0 as a repeated root, so y1=y2y_{1} = y_{2} = 0 and y1⋅y2y_{1} \cdot y_{2} = 0.

Why the others are wrong

  • AThis is the value of x² or the radius squared of the circle, not the product of the y-coordinates.
  • CThis results from finding only one solution and not computing the product.
  • DThis results from a sign error when interpreting the equation.
Question 11Medium

x2+y2=61x^{2} + y^{2} = 61 x+y=11x + y = 11 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1⋅x2x_{1} \cdot x_{2}?

Show the answer and explanation

Why A is right

From the second equation, y=11−xy = 11 - x. Substituting into the circle equation gives x2x^{2} + (11−x)2=61(11 - x)^{2} = 61, which expands to x2x^{2} + 121−22x121 - 22x + x2x^{2} = 61, or 2x2−22x+60=02x^{2} - 22x + 60 = 0. Dividing by 2 gives x2x^{2} - 11x+30=011x + 30 = 0. By Vieta's formulas, the product of the roots is 30, so x1⋅x2x_{1} \cdot x_{2} = 30.

Why the others are wrong

  • BThis results from a sign error when applying Vieta's formulas to the product.
  • CThis represents finding only x = 5 as one solution without recognizing the second solution.
  • DThis results from confusing the sum of the roots with their product.
Question 12Medium

y=−x2+7x−6y = - x^{2} + 7x - 6 y=x−2y = x - 2 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x12⋅x22x_{1} ^{2} \cdot x _{2} ^{2}?

Show the answer and explanation

Why A is right

Substituting y=x−2y = x - 2 into the first equation yields x−2x - 2 = -x2x^{2} + 7x−67x - 6, which simplifies to x2x^{2} - 6x+4=06x + 4 = 0. The product of roots is 4. Since x12x_{1} ^{2} · x22x_{2} ^{2} = (x1x2)2(x_{1} x_{2}) ^{2}, we get (4)2=16(4)^{2} = 16.

Why the others are wrong

  • BThis is the sum x₁ + x₂ raised to some power, not (x₁x₂)².
  • CThis is the product x₁x₂, not its square.
  • DThis is the negative of the correct value, resulting from a sign error in calculation.
Question 13Medium

y=−x2+9y = - x^{2} + 9 y=2x+6y = 2x + 6 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1+x2x_{1} + x_{2}?

Show the answer and explanation

Why A is right

Substituting y=2x+6y = 2x + 6 into the first equation yields 2x+62x + 6 = -x2x^{2} + 9. Rearranging gives x2x^{2} + 2x−3=02x - 3 = 0. By Vieta's formulas, the sum of the roots is -2/1 = -2.

Why the others are wrong

  • BThis results from incorrectly taking the coefficient of x without the proper sign change.
  • CThis is the sum of the y-coordinates, y₁ + y₂, not the x-coordinates.
  • DThis results from finding only one solution instead of summing both solutions.
Question 14Medium

y=x2−2x−15y = x^{2} - 2x - 15 y=4x−3y = 4x - 3 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1+y2y_{1} + y_{2}?

Show the answer and explanation

Why A is right

Substituting y=4x−3y = 4x - 3 into the first equation gives 4x−34x - 3 = x2x^{2} - 2x−152x - 15, which simplifies to x2x^{2} - 6x−12=06x - 12 = 0. By Vieta's formulas, x1+x2x_{1} + x_{2} = 6. Since y1y_{1} = 4x1x_{1} - 3 and y2y_{2} = 4x2x_{2} - 3, we have y1+y2y_{1} + y_{2} = 4(x1+x2)4(x_{1} + x_{2}) - 6 = 4(6) - 6 = 18.

Why the others are wrong

  • BThis results from a sign error in the final calculation.
  • CThis is the sum of the x-coordinates, not the y-coordinates.
  • DThis represents only one y-coordinate value, not their sum.
Question 15Medium

y=−x2+3x+10y = - x^{2} + 3x + 10 y=2x+4y = 2x + 4 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1x2x_{1} x_{2}?

Show the answer and explanation

Why C is right

Substituting y=2x+4y = 2x + 4 into the first equation gives 2x+42x + 4 = -x2x^{2} + 3x+103x + 10. Rearranging yields x2x^{2} - x−6=0x - 6 = 0. By Vieta's formulas, the product of the roots is -6.

Why the others are wrong

  • AThis results from confusing the product of y-coordinates with the product of x-coordinates.
  • BThis results from a sign error when applying Vieta's formulas or rearranging the equation.
  • DThis results from finding only one solution x = 2 and using it as the answer.
Question 16Medium

x2+y2=50x^{2} + y^{2} = 50 x=y+2x = y + 2 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

Show the answer and explanation

Why A is right

Substituting x=y+2x = y + 2 into x2x^{2} + y2y^{2} = 50 yields (y+2)2(y + 2)^{2} + y2y^{2} = 50. Expanding gives y2y^{2} + 4y+44y + 4 + y2y^{2} = 50, which simplifies to 2y2+4y−46=02y^{2} + 4y - 46 = 0 or y2y^{2} + 2y−23=02y - 23 = 0. By Vieta's formulas, y1⋅y2y_{1} \cdot y_{2} = -23.

Why the others are wrong

  • BThis results from a sign error when applying Vieta's formulas to the constant term.
  • CThis results from computing x₁ · x₂ instead of y₁ · y₂, or using only part of Vieta's formulas.
  • DThis is the value of one of the y-coordinates only, not the product of both.
Question 17Medium

x2+y2=40x^{2} + y^{2} = 40 x−y=2x - y = 2 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1⋅y2y_{1} \cdot y_{2}?

Show the answer and explanation

Why A is right

From x−y=2x - y = 2, we get x=y+2x = y + 2. Substituting into x2x^{2} + y2y^{2} = 40 yields (y+2)2(y + 2)^{2} + y2y^{2} = 40, which expands to 2y2+4y+4=402y^{2} + 4y + 4 = 40, or 2y2+4y−36=02y^{2} + 4y - 36 = 0. Dividing by 2 gives y2y^{2} + 2y−18=02y - 18 = 0. By Vieta's formulas, the product of the roots is c/a=−18/1=−18c/a = -18/1 = -18.

Why the others are wrong

  • BThis results from an incorrect sign when applying Vieta's formulas for the product.
  • CThis is y₁ alone (one solution), not the product y₁ · y₂.
  • DThis results from finding x₁ · x₂ instead of y₁ · y₂.
Question 18Medium

y=x2−2x−3y = x^{2} - 2x - 3 y=x+5y = x + 5 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1+y2y_{1} + y_{2}?

Show the answer and explanation

Why B is right

Substituting y=x+5y = x + 5 into the first equation yields x+5x + 5 = x2x^{2} - 2x−32x - 3, which simplifies to x2x^{2} - 3x−8=03x - 8 = 0. The sum of the roots is 3. Since y=x+5y = x + 5, we have y1+y2y_{1} + y_{2} = (x1+5)(x_{1} + 5) + (x2x_{2} + 5) = (x1+x2)(x_{1} + x_{2}) + 10 = 3 + 10 = 13.

Why the others are wrong

  • AThis incorrectly uses only the sum of x-coordinates (x₁ + x₂ = 3) without the transformation to y.
  • CThis is the value of one y-coordinate (y = 4 + 5 = 9), not the sum of both.
  • DThis results from a sign error in computing the sum.
Question 19Medium

y=x2+2x−15y = x^{2} + 2x - 15 y=4x−3y = 4x - 3 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of x1x2x_{1} x_{2}?

Show the answer and explanation

Why A is right

Substituting y=4x−3y = 4x - 3 into the first equation gives 4x−34x - 3 = x2x^{2} + 2x−152x - 15, which simplifies to x2x^{2} - 2x−12=02x - 12 = 0. By Vieta's formulas, the product of the roots equals the constant term divided by the coefficient of x2x^{2}, giving x1x2x_{1} x_{2} = -12.

Why the others are wrong

  • BThis results from a sign error when applying Vieta's formulas, taking the absolute value of -12.
  • CThis results from incorrectly calculating the sum of the x-coordinates instead of the product.
  • DThis results from finding only x = 3 as a solution and not recognizing there is a second solution.
Question 20Medium

y=x2−7x+10y = x^{2} - 7x + 10 y=3x−8y = 3x - 8 The system of equations above has solutions (x1,y1)(x_{1}, y_{1}) and (x2,y2)(x_{2}, y_{2}). What is the value of y1+y2y_{1} + y_{2}?

Show the answer and explanation

Why D is right

Substituting y=3x−8y = 3x - 8 into the first equation gives 3x−83x - 8 = x2x^{2} - 7x+107x + 10, which simplifies to x2x^{2} - 10x+18=010x + 18 = 0. By Vieta's formulas, x1+x2x_{1} + x_{2} = 10. Since y=3x−8y = 3x - 8, we have y1+y2y_{1} + y_{2} = 3(x1+x2)3(x_{1} + x_{2}) - 16 = 3(10) - 16 = 30 - 16 = 14.

Why the others are wrong

  • AThis is the value of x₁ + x₂ without applying the transformation to get y₁ + y₂.
  • BThis results from a sign error in the calculation.
  • CThis is one of the individual y-values from solving the system, not the sum.

What to do after medium

Medium is the tier that decides most scores. If these are landing, the hard set is where the remaining points are.

Easy Systems with nonlinear equations questionsHard Systems with nonlinear equations questionsStart free →

More on Systems with nonlinear equations

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.