Question

use elimination to solve each system of equations.

1. $-v + w = 7$

$v + w = 1$
solution (_, _)

1. $y + z = 4$

$y - z = 8$
solution (_, _)

1. $-4x + 5y = 17$

$4x + 6y = -6$
solution (_, _)

1. $5m - 2p = 24$

$3m + 2p = 24$
solution (_, _)

1. $3j + 4k = 23.5$

$8j - 4k = 4$
solution (_, _)

1. $-7r - 8t = -21$

$7r + 5t = 21$
solution (_, _)
test review
solve for x.
a. $\frac{x}{4} - 12 = 4$
b. $\frac{7}{8}x = 14$

Explanation:

Problem 1: Eliminate \(v\)

Add the two equations:
$$(-v + w) + (v + w) = 7 + 1$$
$$2w = 8$$

Problem 1: Solve for \(w\)

$$w = \frac{8}{2} = 4$$

Problem 1: Solve for \(v\)

Substitute \(w=4\) into \(v + w = 1\):
$$v + 4 = 1 \implies v = 1 - 4 = -3$$

---

Problem 2: Eliminate \(z\)

Add the two equations:
$$(y + z) + (y - z) = 4 + 8$$
$$2y = 12$$

Problem 2: Solve for \(y\)

$$y = \frac{12}{2} = 6$$

Problem 2: Solve for \(z\)

Substitute \(y=6\) into \(y + z = 4\):
$$6 + z = 4 \implies z = 4 - 6 = -2$$

---

Problem 3: Eliminate \(x\)

Add the two equations:
$$(-4x + 5y) + (4x + 6y) = 17 + (-6)$$
$$11y = 11$$

Problem 3: Solve for \(y\)

$$y = \frac{11}{11} = 1$$

Problem 3: Solve for \(x\)

Substitute \(y=1\) into \(4x + 6y = -6\):
$$4x + 6(1) = -6 \implies 4x = -12 \implies x = \frac{-12}{4} = -3$$

---

Problem 4: Eliminate \(p\)

Add the two equations:
$$(5m - 2p) + (3m + 2p) = 24 + 24$$
$$8m = 48$$

Problem 4: Solve for \(m\)

$$m = \frac{48}{8} = 6$$

Problem 4: Solve for \(p\)

Substitute \(m=6\) into \(3m + 2p = 24\):
$$3(6) + 2p = 24 \implies 18 + 2p = 24 \implies 2p = 6 \implies p = 3$$

---

Problem 5: Eliminate \(k\)

Add the two equations:
$$(3j + 4k) + (8j - 4k) = 23.5 + 4$$
$$11j = 27.5$$

Problem 5: Solve for \(j\)

$$j = \frac{27.5}{11} = 2.5$$

Problem 5: Solve for \(k\)

Substitute \(j=2.5\) into \(8j - 4k = 4\):
$$8(2.5) - 4k = 4 \implies 20 - 4k = 4 \implies -4k = -16 \implies k = 4$$

---

Problem 6: Eliminate \(r\)

Add the two equations:
$$(-7r - 8t) + (7r + 5t) = -21 + 21$$
$$-3t = 0$$

Problem 6: Solve for \(t\)

$$t = \frac{0}{-3} = 0$$

Problem 6: Solve for \(r\)

Substitute \(t=0\) into \(7r + 5t = 21\):
$$7r + 5(0) = 21 \implies 7r = 21 \implies r = 3$$

---

Test Review a: Isolate \(\frac{x}{4}\)

Add 12 to both sides:
$$\frac{x}{4} = 4 + 12$$
$$\frac{x}{4} = 16$$

Test Review a: Solve for \(x\)

Multiply both sides by 4:
$$x = 16 \times 4 = 64$$

---

Test Review b: Solve for \(x\)

Multiply both sides by \(\frac{8}{7}\):
$$x = 14 \times \frac{8}{7} = 16$$

Answer:

1. Solution \((-3, 4)\)
2. Solution \((6, -2)\)
3. Solution \((-3, 1)\)
4. Solution \((6, 3)\)
5. Solution \((2.5, 4)\)
6. Solution \((3, 0)\)

Test Review:
a. \(x = 64\)
b. \(x = 16\)