436 FINAL QUESTIONS ON SYSTEMS OF EQUATIONS |

Solve these questions. Solutions at the end of the file.

1) State which of the following systems of lines are **parallel, coincident **or** skew** and state **how many solutions** each system has.

a). y = 2x – 5. y = 2x + 1 |
b). y = 3x + 4. y = 2x – 4 |
c). 3x + 6y = 12. x + 2y = 4 |
d). x – y + 8 = 02 x – 10 = 2y |
e) 2 x + y = 63 x – y = 4 |

2) Solve these systems algebraically:

a) 4 x + y = 32 x – 3y = 12 |
b) 3 x – 5y = 352.5 x + y = 8.5 |
c). y = x^{ 2} – 4x. y = x + 6 |
d). x^{ 2} + y^{ 2} = 5. y = x – 3 |

3)

- a) Find the equation of the line through A( – 4, 3) and B(6, – 2).

b) Write the equation of the line through P(0, 2) with a slope of .

c) Find the point of intersection of the lines in parts a) and b).

4) Write the equation of the **horizontal tangent line** to the parabola *y* = *x*^{ 2} – 2*x* – 3.

5) Viv and Jaclyn sell magazine subscriptions.

Viv's weekly salary is $100 plus a commission of $6.50 per subscription sold.

Jaclyn earns $75 per week plus a $7.00 commission for each subscription sold.

- a) How many subscriptions must they sell in order to earn the same amount?

b) How much do they each earn?

6) Find the points of intersection of the circle *x*^{ 2} + *y*^{ 2} = 125 and the line *y* = – *x* + 15.

7) The perimeter of a rectangle is 138 cm. If the length were 3 cm. longer,

it would be twice the width. What are the dimensions of this rectangle?

8) A group of LCC students went to the greasy spoon for burgers and fries.

5 burgers and 3 fries cost $11.25. 1 burger and 3 fries cost $5.25.

How much would it cost for 7 burgers and 5 fries?

**Solutions**

1) State which of the following systems of lines are **parallel, coincident **or** skew** and state **how many solutions** each system has.

a) // 0 soln |
b) skew 1-soln |
c) coincident, inf |
d) // 0 soln |
e) skew 1-soln |

2) Solve these systems algebraically:

a) 4 x + y = 32 x – 3y = 12substitution for 2 x = 1.5, y = – 3 |
b) 3 x – 5y = 352.5 x + y = 8.5substitution for 3 |
c). y = x^{ 2} – 4x. y = x + 6comparison. . x² – 5x – 6 = 0( x – 6)(x + 1) = 0if x = 6, y = 12if x = – 1, y = 5 |
d). x^{ 2} + y^{ 2} = 5. y = x – 3substitution for |

3)

a) y = – ½ x + 1 |
b) y = x + 2 | c) () |

4) *y = x*² – 2*x *– 3 and we know which means *k* = – 4

since the *y*-value of the vertex is – 4, the equation of the horizontal tangent is ** y = – 4**.

5) a) Let A_{V} = Viv's Amount and let A_{J} = Jaclyn's Amount, and let *x* = # of subscriptions sold.

A_{V} = 100 + 6.50*x* ........ and ......... A_{J} = 75 + 7.00*x*

Since we want their Amount to be the same, we set A_{V} = A_{J} .

100 + 6.50*x* = 75 + 7.00*x* becomes 0.50*x* = 25, so *x* = 50.

They must sell 50 subscriptions to earn the same Amount.

b) Set *x* = 50 in either of the 2 equations to find they earn $425.00 each.

6) if we set *y* = – *x* + 15, then *x*^{ 2} + *y*^{ 2} = 125 becomes *x*^{ }² + (15 – *x*)² = 125

We get *x*^{ }² + 225 – 30*x* + *x*² = 125

So, 2*x*^{ }² – 30*x* + 100 = 0 which is *x*^{ }² – 15*x* + 50 = 0

or (*x* – 10)(*x* – 5) = 0 so the points of intersection are (5, 10) (10, 5)

7)

8) Let *b* = the cost of 1 burger ....... and ....... Let *f* = the cost of 1 order of fries.

5*b* + 3*f* = 11.25 .......and .......*b* + 3*f* = 5.25 which becomes *b* = 5.25 – 3*f* (substitute)

5(5.25 – 3*f*) + 3*f* = 11.25, so 26.25 – 15*f* + 3*f* = 11.25

– 12*f* = –15, ....so ....* f* = $1.25 .......and ....... *b* = $1.50

burger = $1.50, fries = $1.25, 7 burgers + 5 fries = $16.75

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