Question

1. solve for x.

1. find jl.

1. find gh.

1. solve for x.

Explanation:

Problem 5: Solve for x

Step1: Apply Midsegment Theorem

The midsegment of a triangle is half the length of the parallel side, so:
$$8x - 23 = \frac{1}{2}(10x + 44)$$

Step2: Eliminate fraction, multiply by 2

$$2(8x - 23) = 10x + 44$$
$$16x - 46 = 10x + 44$$

Step3: Isolate x terms

$$16x - 10x = 44 + 46$$
$$6x = 90$$

Step4: Solve for x

$$x = \frac{90}{6} = 15$$

Problem 6: Solve for x

Step1: Apply Midsegment Theorem

The midsegment is half the length of the parallel side:
$$6x + 7 = \frac{1}{2}(19x - 28)$$

Step2: Eliminate fraction, multiply by 2

$$2(6x + 7) = 19x - 28$$
$$12x + 14 = 19x - 28$$

Step3: Isolate x terms

$$14 + 28 = 19x - 12x$$
$$42 = 7x$$

Step4: Solve for x

$$x = \frac{42}{7} = 6$$

Problem 7: Find JL

Step1: Apply Midsegment Theorem

$MN$ is the midsegment, so $MN = \frac{1}{2}JL$:
$$5x - 16 = \frac{1}{2}(4x + 34)$$

Step2: Eliminate fraction, multiply by 2

$$2(5x - 16) = 4x + 34$$
$$10x - 32 = 4x + 34$$

Step3: Isolate x terms

$$10x - 4x = 34 + 32$$
$$6x = 66$$

Step4: Solve for x

$$x = \frac{66}{6} = 11$$

Step5: Substitute x into JL expression

$$JL = 4(11) + 34 = 44 + 34 = 78$$

Problem 8: Find GH

Step1: Apply Midsegment Theorem

$GH$ is the midsegment, so $GH = \frac{1}{2}OF$:
$$3x - 4 = \frac{1}{2}(9x - 59)$$

Step2: Eliminate fraction, multiply by 2

$$2(3x - 4) = 9x - 59$$
$$6x - 8 = 9x - 59$$

Step3: Isolate x terms

$$-8 + 59 = 9x - 6x$$
$$51 = 3x$$

Step4: Solve for x

$$x = \frac{51}{3} = 17$$

Step5: Substitute x into GH expression

$$GH = 3(17) - 4 = 51 - 4 = 47$$

Answer:

1. Problem 5: $x = 15$
2. Problem 6: $x = 6$
3. Problem 7: $JL = 78$
4. Problem 8: $GH = 47$