Multi-Step Equations Exercises
Multi-step equations practice problems with answers.
For this exercise, I have prepared seven (7) multi-step equations for you to practice. If you feel the need to review the techniques involved in solving multi-step equations, take a short detour to review my other lesson about it. Click the link below to take you there!
Solving Multi-Step Equations
1) Solve the multi-step equation for [latex]\large{c}[/latex].
[latex]c – 20 = 4 – 3c[/latex]
Add both sides by [latex]20[/latex]. Next, add [latex]3c[/latex] to both sides. Finally, divide both sides by the coefficient of [latex]4c[/latex] which is [latex]4[/latex] to get [latex]c=6[/latex].
2) Solve the multi-step equation for [latex]\large{n}[/latex].
[latex] – \,4\left( { – 3n – 8} \right) = 10n + 20[/latex]
- Remember to always perform the same operation on both sides of the equation.
- Subtract by [latex]32[/latex].
- Subtract by [latex]10n[/latex].
- Divide by [latex]2[/latex]
- The final solution is [latex]n=-6[/latex].
3) Solve the multi-step equation for [latex]\large{y}[/latex].
[latex]2\left( {4 – y} \right) – 3\left( {y + 3} \right) = – 11[/latex]
Apply twice the Distributive Property of Multiplication over Addition to the left side of the equation. Then combine like terms . Add both sides by [latex]1[/latex] followed by dividing both sides of the equation by [latex]-5[/latex].
4) Solve the multi-step equation for [latex]\large{k}[/latex].
[latex]{\Large{{6k + 4} \over 2}} = 2k – 11[/latex]
Multiply both sides by [latex]2[/latex]. Next, subtract [latex]4[/latex] to both sides. Then, subtract [latex]4k[/latex]. Finally, divide by [latex]2[/latex] to obtain the value of [latex]k[/latex] which is [latex]-13[/latex].
5) Solve the multi-step equation for [latex]\large{x}[/latex].
[latex] – \left( { – 8 – 3x} \right) = – 2\left( {1 – x} \right) + 6x[/latex]
Apply the Distributive Property on both sides of the equation. Be careful when multiplying expressions with the same or different signs . Next, add [latex]2[/latex] to both sides, then subtract [latex]3x[/latex], and finally finish it off by dividing [latex]5[/latex] to both sides.
6) Solve the multi-step equation for [latex]\large{m}[/latex].
[latex]{\large{3 \over 4}}m – 2\left( {m – 1} \right) = {\large{1 \over 4}}m + 5[/latex]
7) Solve the multi-step equation for [latex]\large{x}[/latex].
[latex]3\left( {3x – 8} \right) – 5\left( {3x – 8} \right) = 4\left( {x – 2} \right) – 6\left( {x – 2} \right)[/latex]
You might also like these tutorials:
- Two-Step Equations Practice Problems with Answers
9.1 Verifying Trigonometric Identities and Using Trigonometric Identities to Simplify Trigonometric Expressions
sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ sin 2 θ − 1 tan θ sin θ − tan θ = ( sin θ + 1 ) ( sin θ − 1 ) tan θ ( sin θ − 1 ) = sin θ + 1 tan θ
This is a difference of squares formula: 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) . 25 − 9 sin 2 θ = ( 5 − 3 sin θ ) ( 5 + 3 sin θ ) .
9.2 Sum and Difference Identities
2 + 6 4 2 + 6 4
2 − 6 4 2 − 6 4
1 − 3 1 + 3 1 − 3 1 + 3
cos ( 5 π 14 ) cos ( 5 π 14 )
9.3 Double-Angle, Half-Angle, and Reduction Formulas
cos ( 2 α ) = 7 32 cos ( 2 α ) = 7 32
cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ ) cos 4 θ − sin 4 θ = ( cos 2 θ + sin 2 θ ) ( cos 2 θ − sin 2 θ ) = cos ( 2 θ )
cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ cos ( 2 θ ) cos θ = ( cos 2 θ − sin 2 θ ) cos θ = cos 3 θ − cos θ sin 2 θ
10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x ) 10 cos 4 x = 10 ( cos 2 x ) 2 = 10 [ 1 + cos ( 2 x ) 2 ] 2 Substitute reduction formula for cos 2 x . = 10 4 [ 1 + 2 cos ( 2 x ) + cos 2 ( 2 x ) ] = 10 4 + 10 2 cos ( 2 x ) + 10 4 ( 1 + cos 2 ( 2 x ) 2 ) Substitute reduction formula for cos 2 x . = 10 4 + 10 2 cos ( 2 x ) + 10 8 + 10 8 cos ( 4 x ) = 30 8 + 5 cos ( 2 x ) + 10 8 cos ( 4 x ) = 15 4 + 5 cos ( 2 x ) + 5 4 cos ( 4 x )
− 2 5 − 2 5
9.4 Sum-to-Product and Product-to-Sum Formulas
1 2 ( cos 6 θ + cos 2 θ ) 1 2 ( cos 6 θ + cos 2 θ )
1 2 ( sin 2 x + sin 2 y ) 1 2 ( sin 2 x + sin 2 y )
− 2 − 3 4 − 2 − 3 4
2 sin ( 2 θ ) cos ( θ ) 2 sin ( 2 θ ) cos ( θ )
tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ tan θ cot θ − cos 2 θ = ( sin θ cos θ ) ( cos θ sin θ ) − cos 2 θ = 1 − cos 2 θ = sin 2 θ
9.5 Solving Trigonometric Equations
x = 7 π 6 , 11 π 6 x = 7 π 6 , 11 π 6
π 3 ± π k π 3 ± π k
θ ≈ 1.7722 ± 2 π k θ ≈ 1.7722 ± 2 π k and θ ≈ 4.5110 ± 2 π k θ ≈ 4.5110 ± 2 π k
cos θ = − 1 , θ = π cos θ = − 1 , θ = π
π 2 , 2 π 3 , 4 π 3 , 3 π 2 π 2 , 2 π 3 , 4 π 3 , 3 π 2
9.1 Section Exercises
All three functions, F F , G G , and H H , are even.
This is because F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) F ( − x ) = sin ( − x ) sin ( − x ) = ( − sin x ) ( − sin x ) = sin 2 x = F ( x ) , G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) G ( − x ) = cos ( − x ) cos ( − x ) = cos x cos x = cos 2 x = G ( x ) and H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) . H ( − x ) = tan ( − x ) tan ( − x ) = ( − tan x ) ( − tan x ) = tan 2 x = H ( x ) .
When cos t = 0 , cos t = 0 , then sec t = 1 0 , sec t = 1 0 , which is undefined.
sin x sin x
sec x sec x
csc t csc t
sec 2 x sec 2 x
sin 2 x + 1 sin 2 x + 1
1 sin x 1 sin x
1 cot x 1 cot x
tan x tan x
− 4 sec x tan x − 4 sec x tan x
± 1 cot 2 x + 1 ± 1 cot 2 x + 1
± 1 − sin 2 x sin x ± 1 − sin 2 x sin x
Answers will vary. Sample proof:
cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x cos x − cos 3 x = cos x ( 1 − cos 2 x ) = cos x sin 2 x
Answers will vary. Sample proof: 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x 1 + sin 2 x cos 2 x = 1 cos 2 x + sin 2 x cos 2 x = sec 2 x + tan 2 x = tan 2 x + 1 + tan 2 x = 1 + 2 tan 2 x
Answers will vary. Sample proof: cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x cos 2 x − tan 2 x = 1 − sin 2 x − ( sec 2 x − 1 ) = 1 − sin 2 x − sec 2 x + 1 = 2 − sin 2 x − sec 2 x
Proved with negative and Pythagorean identities
True 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ 3 sin 2 θ + 4 cos 2 θ = 3 sin 2 θ + 3 cos 2 θ + cos 2 θ = 3 ( sin 2 θ + cos 2 θ ) + cos 2 θ = 3 + cos 2 θ
9.2 Section Exercises
The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x , x , the second angle measures π 2 − x . π 2 − x . Then sin x = cos ( π 2 − x ) . sin x = cos ( π 2 − x ) . The same holds for the other cofunction identities. The key is that the angles are complementary.
sin ( − x ) = − sin x , sin ( − x ) = − sin x , so sin x sin x is odd. cos ( − x ) = cos ( 0 − x ) = cos x , cos ( − x ) = cos ( 0 − x ) = cos x , so cos x cos x is even.
6 − 2 4 6 − 2 4
− 2 − 3 − 2 − 3
− 2 2 sin x − 2 2 cos x − 2 2 sin x − 2 2 cos x
− 1 2 cos x − 3 2 sin x − 1 2 cos x − 3 2 sin x
csc θ csc θ
cot x cot x
tan ( x 10 ) tan ( x 10 )
sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15 sin ( a − b ) = ( 4 5 ) ( 1 3 ) − ( 3 5 ) ( 2 2 3 ) = 4 − 6 2 15 cos ( a + b ) = ( 3 5 ) ( 1 3 ) − ( 4 5 ) ( 2 2 3 ) = 3 − 8 2 15
cot ( π 6 − x ) cot ( π 6 − x )
cot ( π 4 + x ) cot ( π 4 + x )
sin x 2 + cos x 2 sin x 2 + cos x 2
They are the same.
They are the different, try g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) . g ( x ) = sin ( 9 x ) − cos ( 3 x ) sin ( 6 x ) .
They are the different, try g ( θ ) = 2 tan θ 1 − tan 2 θ . g ( θ ) = 2 tan θ 1 − tan 2 θ .
They are different, try g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) . g ( x ) = tan x − tan ( 2 x ) 1 + tan x tan ( 2 x ) .
− 3 − 1 2 2 , or − 0.2588 − 3 − 1 2 2 , or − 0.2588
1 + 3 2 2 , 1 + 3 2 2 , or 0.9659
tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x tan ( x + π 4 ) = tan x + tan ( π 4 ) 1 − tan x tan ( π 4 ) = tan x + 1 1 − tan x ( 1 ) = tan x + 1 1 − tan x
cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b cos ( a + b ) cos a cos b = cos a cos b cos a cos b − sin a sin b cos a cos b = 1 − tan a tan b
cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h cos ( x + h ) − cos x h = cos x cosh − sin x sinh − cos x h = cos x ( cosh − 1 ) − sin x sinh h = cos x cos h − 1 h − sin x sin h h
True. Note that sin ( α + β ) = sin ( π − γ ) sin ( α + β ) = sin ( π − γ ) and expand the right hand side.
9.3 Section Exercises
Use the Pythagorean identities and isolate the squared term.
1 − cos x sin x , sin x 1 + cos x , 1 − cos x sin x , sin x 1 + cos x , multiplying the top and bottom by 1 − cos x 1 − cos x and 1 + cos x , 1 + cos x , respectively.
a) 3 7 32 3 7 32 b) 31 32 31 32 c) 3 7 31 3 7 31
a) 3 2 3 2 b) − 1 2 − 1 2 c) − 3 − 3
cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2 cos θ = − 2 5 5 , sin θ = 5 5 , tan θ = − 1 2 , csc θ = 5 , sec θ = − 5 2 , cot θ = − 2
2 sin ( π 2 ) 2 sin ( π 2 )
2 − 2 2 2 − 2 2
2 − 3 2 2 − 3 2
2 + 3 2 + 3
− 1 − 2 − 1 − 2
a) 3 13 13 3 13 13 b) − 2 13 13 − 2 13 13 c) − 3 2 − 3 2
a) 10 4 10 4 b) 6 4 6 4 c) 15 3 15 3
120 169 , – 119 169 , – 120 119 120 169 , – 119 169 , – 120 119
2 13 13 , 3 13 13 , 2 3 2 13 13 , 3 13 13 , 2 3
cos ( 74° ) cos ( 74° )
cos ( 18 x ) cos ( 18 x )
3 sin ( 10 x ) 3 sin ( 10 x )
− 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x ) − 2 sin ( − x ) cos ( − x ) = − 2 ( − sin ( x ) cos ( x ) ) = sin ( 2 x )
sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ sin ( 2 θ ) 1 + cos ( 2 θ ) tan 2 θ = 2 sin ( θ ) cos ( θ ) 1 + cos 2 θ − sin 2 θ tan 2 θ = 2 sin ( θ ) cos ( θ ) 2 cos 2 θ tan 2 θ = sin ( θ ) cos θ tan 2 θ = cot ( θ ) tan 2 θ = tan 3 θ
1 + cos ( 12 x ) 2 1 + cos ( 12 x ) 2
3 + cos ( 12 x ) − 4 cos ( 6 x ) 8 3 + cos ( 12 x ) − 4 cos ( 6 x ) 8
2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32 2 + cos ( 2 x ) − 2 cos ( 4 x ) − cos ( 6 x ) 32
3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 3 + cos ( 4 x ) + 4 cos ( 2 x )
1 − cos ( 4 x ) 8 1 − cos ( 4 x ) 8
3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 ) 3 + cos ( 4 x ) − 4 cos ( 2 x ) 4 ( cos ( 2 x ) + 1 )
( 1 + cos ( 4 x ) ) sin x 2 ( 1 + cos ( 4 x ) ) sin x 2
4 sin x cos x ( cos 2 x − sin 2 x ) 4 sin x cos x ( cos 2 x − sin 2 x )
2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x ) 2 tan x 1 + tan 2 x = 2 sin x cos x 1 + sin 2 x cos 2 x = 2 sin x cos x cos 2 x + sin 2 x cos 2 x = 2 sin x cos x . cos 2 x 1 = 2 sin x cos x = sin ( 2 x )
2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x ) 2 sin x cos x 2 cos 2 x − 1 = sin ( 2 x ) cos ( 2 x ) = tan ( 2 x )
sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x sin ( x + 2 x ) = sin x cos ( 2 x ) + sin ( 2 x ) cos x = sin x ( cos 2 x − sin 2 x ) + 2 sin x cos x cos x = sin x cos 2 x − sin 3 x + 2 sin x cos 2 x = 3 sin x cos 2 x − sin 3 x
1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1 1 + cos ( 2 t ) sin ( 2 t ) − cos t = 1 + 2 cos 2 t − 1 2 sin t cos t − cos t = 2 cos 2 t cos t ( 2 sin t − 1 ) = 2 cos t 2 sin t − 1
( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) − sin ( 8 x ) ) ( cos 2 ( 4 x ) − sin 2 ( 4 x ) + sin ( 8 x ) ) = = ( cos ( 8 x ) − sin ( 8 x ) ) ( cos ( 8 x ) + sin ( 8 x ) ) = cos 2 ( 8 x ) − sin 2 ( 8 x ) = cos ( 16 x )
9.4 Section Exercises
Substitute α α into cosine and β β into sine and evaluate.
Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: sin ( 3 x ) + sin x cos x = 1. sin ( 3 x ) + sin x cos x = 1. When converting the numerator to a product the equation becomes: 2 sin ( 2 x ) cos x cos x = 1 2 sin ( 2 x ) cos x cos x = 1
8 ( cos ( 5 x ) − cos ( 27 x ) ) 8 ( cos ( 5 x ) − cos ( 27 x ) )
sin ( 2 x ) + sin ( 8 x ) sin ( 2 x ) + sin ( 8 x )
1 2 ( cos ( 6 x ) − cos ( 4 x ) ) 1 2 ( cos ( 6 x ) − cos ( 4 x ) )
2 cos ( 5 t ) cos t 2 cos ( 5 t ) cos t
2 cos ( 7 x ) 2 cos ( 7 x )
2 cos ( 6 x ) cos ( 3 x ) 2 cos ( 6 x ) cos ( 3 x )
1 4 ( 1 + 3 ) 1 4 ( 1 + 3 )
1 4 ( 3 − 2 ) 1 4 ( 3 − 2 )
1 4 ( 3 − 1 ) 1 4 ( 3 − 1 )
cos ( 80° ) − cos ( 120° ) cos ( 80° ) − cos ( 120° )
1 2 ( sin ( 221° ) + sin ( 205° ) ) 1 2 ( sin ( 221° ) + sin ( 205° ) )
2 cos ( 31° ) 2 cos ( 31° )
2 cos ( 66.5° ) sin ( 34.5° ) 2 cos ( 66.5° ) sin ( 34.5° )
2 sin ( −1.5° ) cos ( 0.5° ) 2 sin ( −1.5° ) cos ( 0.5° )
2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x ) 2 sin ( 7 x ) − 2 sin x = 2 sin ( 4 x + 3 x ) − 2 sin ( 4 x − 3 x ) = 2 ( sin ( 4 x ) cos ( 3 x ) + sin ( 3 x ) cos ( 4 x ) ) − 2 ( sin ( 4 x ) cos ( 3 x ) − sin ( 3 x ) cos ( 4 x ) ) = 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) − 2 sin ( 4 x ) cos ( 3 x ) + 2 sin ( 3 x ) cos ( 4 x ) ) = 4 sin ( 3 x ) cos ( 4 x )
sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x sin x + sin ( 3 x ) = 2 sin ( 4 x 2 ) cos ( − 2 x 2 ) = 2 sin ( 2 x ) cos x = 2 ( 2 sin x cos x ) cos x = 4 sin x cos 2 x
2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) ) 2 tan x cos ( 3 x ) = 2 sin x cos ( 3 x ) cos x = 2 ( .5 ( sin ( 4 x ) − sin ( 2 x ) ) ) cos x = 1 cos x ( sin ( 4 x ) − sin ( 2 x ) ) = sec x ( sin ( 4 x ) − sin ( 2 x ) )
2 cos ( 35° ) cos ( 23° ) , 1.5081 2 cos ( 35° ) cos ( 23° ) , 1.5081
− 2 sin ( 33° ) sin ( 11° ) , − 0.2078 − 2 sin ( 33° ) sin ( 11° ) , − 0.2078
1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410 1 2 ( cos ( 99° ) − cos ( 71° ) ) , −0.2410
It is an identity.
It is not an identity, but 2 cos 3 x 2 cos 3 x is.
tan ( 3 t ) tan ( 3 t )
2 cos ( 2 x ) 2 cos ( 2 x )
− sin ( 14 x ) − sin ( 14 x )
Start with cos x + cos y . cos x + cos y . Make a substitution and let x = α + β x = α + β and let y = α − β , y = α − β , so cos x + cos y cos x + cos y becomes cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β cos ( α + β ) + cos ( α − β ) = cos α cos β − sin α sin β + cos α cos β + sin α sin β = 2 cos α cos β
Since x = α + β x = α + β and y = α − β , y = α − β , we can solve for α α and β β in terms of x and y and substitute in for 2 cos α cos β 2 cos α cos β and get 2 cos ( x + y 2 ) cos ( x − y 2 ) . 2 cos ( x + y 2 ) cos ( x − y 2 ) .
cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x cos ( 3 x ) + cos x cos ( 3 x ) − cos x = 2 cos ( 2 x ) cos x − 2 sin ( 2 x ) sin x = − cot ( 2 x ) cot x
cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y cos ( 2 y ) − cos ( 4 y ) sin ( 2 y ) + sin ( 4 y ) = − 2 sin ( 3 y ) sin ( − y ) 2 sin ( 3 y ) cos y = 2 sin ( 3 y ) sin ( y ) 2 sin ( 3 y ) cos y = tan y
cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x cos x − cos ( 3 x ) = − 2 sin ( 2 x ) sin ( − x ) = 2 ( 2 sin x cos x ) sin x = 4 sin 2 x cos x
tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t tan ( π 4 − t ) = tan ( π 4 ) − tan t 1 + tan ( π 4 ) tan ( t ) = 1 − tan t 1 + tan t
9.5 Section Exercises
There will not always be solutions to trigonometric function equations. For a basic example, cos ( x ) = −5. cos ( x ) = −5.
If the sine or cosine function has a coefficient of one, isolate the term on one side of the equals sign. If the number it is set equal to has an absolute value less than or equal to one, the equation has solutions, otherwise it does not. If the sine or cosine does not have a coefficient equal to one, still isolate the term but then divide both sides of the equation by the leading coefficient. Then, if the number it is set equal to has an absolute value greater than one, the equation has no solution.
π 3 , 2 π 3 π 3 , 2 π 3
3 π 4 , 5 π 4 3 π 4 , 5 π 4
π 4 , 5 π 4 π 4 , 5 π 4
π 4 , 3 π 4 , 5 π 4 , 7 π 4 π 4 , 3 π 4 , 5 π 4 , 7 π 4
π 4 , 7 π 4 π 4 , 7 π 4
7 π 6 , 11 π 6 7 π 6 , 11 π 6
π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18 π 18 , 5 π 18 , 13 π 18 , 17 π 18 , 25 π 18 , 29 π 18
3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12 3 π 12 , 5 π 12 , 11 π 12 , 13 π 12 , 19 π 12 , 21 π 12
1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6 1 6 , 5 6 , 13 6 , 17 6 , 25 6 , 29 6 , 37 6
0 , π 3 , π , 5 π 3 0 , π 3 , π , 5 π 3
π 3 , π , 5 π 3 π 3 , π , 5 π 3
π 3 , 3 π 2 , 5 π 3 π 3 , 3 π 2 , 5 π 3
0 , π 0 , π
π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 ) π − sin − 1 ( − 1 4 ) , 7 π 6 , 11 π 6 , 2 π + sin − 1 ( − 1 4 )
1 3 ( sin − 1 ( 9 10 ) ) 1 3 ( sin − 1 ( 9 10 ) ) , π 3 − 1 3 ( sin − 1 ( 9 10 ) ) π 3 − 1 3 ( sin − 1 ( 9 10 ) ) , 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 2 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , π − 1 3 ( sin − 1 ( 9 10 ) ) π − 1 3 ( sin − 1 ( 9 10 ) ) , 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) 4 π 3 + 1 3 ( sin − 1 ( 9 10 ) ) , 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) ) 5 π 3 − 1 3 ( sin − 1 ( 9 10 ) )
π 6 , 5 π 6 , 7 π 6 , 11 π 6 π 6 , 5 π 6 , 7 π 6 , 11 π 6
3 π 2 , π 6 , 5 π 6 3 π 2 , π 6 , 5 π 6
0 , π 3 , π , 4 π 3 0 , π 3 , π , 4 π 3
There are no solutions.
cos − 1 ( 1 3 ( 1 − 7 ) ) cos − 1 ( 1 3 ( 1 − 7 ) ) , 2 π − cos − 1 ( 1 3 ( 1 − 7 ) ) 2 π − cos − 1 ( 1 3 ( 1 − 7 ) )
tan − 1 ( 1 2 ( 29 − 5 ) ) tan − 1 ( 1 2 ( 29 − 5 ) ) , π + tan − 1 ( 1 2 ( − 29 − 5 ) ) π + tan − 1 ( 1 2 ( − 29 − 5 ) ) , π + tan − 1 ( 1 2 ( 29 − 5 ) ) π + tan − 1 ( 1 2 ( 29 − 5 ) ) , 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) ) 2 π + tan − 1 ( 1 2 ( − 29 − 5 ) )
0 , 2 π 3 , 4 π 3 0 , 2 π 3 , 4 π 3
sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2 sin − 1 ( 3 5 ) , π 2 , π − sin − 1 ( 3 5 ) , 3 π 2
cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 ) cos − 1 ( − 1 4 ) , 2 π − cos − 1 ( − 1 4 )
π 3 π 3 , cos − 1 ( − 3 4 ) cos − 1 ( − 3 4 ) , 2 π − cos − 1 ( − 3 4 ) 2 π − cos − 1 ( − 3 4 ) , 5 π 3 5 π 3
cos − 1 ( 3 4 ) cos − 1 ( 3 4 ) , cos − 1 ( − 2 3 ) cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( − 2 3 ) 2 π − cos − 1 ( − 2 3 ) , 2 π − cos − 1 ( 3 4 ) 2 π − cos − 1 ( 3 4 )
0 , π 2 , π , 3 π 2 0 , π 2 , π , 3 π 2
π 3 π 3 , cos −1 ( − 1 4 ) cos −1 ( − 1 4 ) , 2 π − cos −1 ( − 1 4 ) 2 π − cos −1 ( − 1 4 ) , 5 π 3 5 π 3
π + tan −1 ( −2 ) π + tan −1 ( −2 ) , π + tan −1 ( − 3 2 ) π + tan −1 ( − 3 2 ) , 2 π + tan −1 ( −2 ) 2 π + tan −1 ( −2 ) , 2 π + tan −1 ( − 3 2 ) 2 π + tan −1 ( − 3 2 )
2 π k + 0.2734 , 2 π k + 2.8682 2 π k + 0.2734 , 2 π k + 2.8682
π k − 0.3277 π k − 0.3277
0.6694 , 1.8287 , 3.8110 , 4.9703 0.6694 , 1.8287 , 3.8110 , 4.9703
1.0472 , 3.1416 , 5.2360 1.0472 , 3.1416 , 5.2360
0.5326 , 1.7648 , 3.6742 , 4.9064 0.5326 , 1.7648 , 3.6742 , 4.9064
sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2 sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) , 3 π 2
π 2 , 3 π 2 π 2 , 3 π 2
7.2 ∘ 7.2 ∘
5.7 ∘ 5.7 ∘
82.4 ∘ 82.4 ∘
31.0 ∘ 31.0 ∘
88.7 ∘ 88.7 ∘
59.0 ∘ 59.0 ∘
36.9 ∘ 36.9 ∘
Review Exercises
sin − 1 ( 3 3 ) sin − 1 ( 3 3 ) , π − sin − 1 ( 3 3 ) π − sin − 1 ( 3 3 ) , π + sin − 1 ( 3 3 ) π + sin − 1 ( 3 3 ) , 2 π − sin − 1 ( 3 3 ) 2 π − sin − 1 ( 3 3 )
sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 ) sin − 1 ( 1 4 ) , π − sin − 1 ( 1 4 )
cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x cos ( 4 x ) − cos ( 3 x ) cos x = cos ( 2 x + 2 x ) − cos ( x + 2 x ) cos x = cos ( 2 x ) cos ( 2 x ) − sin ( 2 x ) sin ( 2 x ) − cos x cos ( 2 x ) cos x + sin x sin ( 2 x ) cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + sin x ( 2 ) sin x cos x cos x = ( cos 2 x − sin 2 x ) 2 − 4 cos 2 x sin 2 x − cos 2 x ( cos 2 x − sin 2 x ) + 2 sin 2 x cos 2 x = cos 4 x − 2 cos 2 x sin 2 x + sin 4 x − 4 cos 2 x sin 2 x − cos 4 x + cos 2 x sin 2 x + 2 sin 2 x cos 2 x = sin 4 x − 4 cos 2 x sin 2 x + cos 2 x sin 2 x = sin 2 x ( sin 2 x + cos 2 x ) − 4 cos 2 x sin 2 x = sin 2 x − 4 cos 2 x sin 2 x
tan ( 5 8 x ) tan ( 5 8 x )
− 24 25 , − 7 25 , 24 7 − 24 25 , − 7 25 , 24 7
2 ( 2 + 2 ) 2 ( 2 + 2 )
2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4 2 10 , 7 2 10 , 1 7 , 3 5 , 4 5 , 3 4
cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x ) = cot x − cos x sin x ( 2 ) sin 2 x = − 2 sin x cos x + cot x = − sin ( 2 x ) + cot x cot x cos ( 2 x ) = cot x ( 1 − 2 sin 2 x ) = cot x − cos x sin x ( 2 ) sin 2 x = − 2 sin x cos x + cot x = − sin ( 2 x ) + cot x
10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 ) 10 sin x − 5 sin ( 3 x ) + sin ( 5 x ) 8 ( cos ( 2 x ) + 1 )
− 2 2 − 2 2
1 2 ( sin ( 6 x ) + sin ( 12 x ) ) 1 2 ( sin ( 6 x ) + sin ( 12 x ) )
2 sin ( 13 2 x ) cos ( 9 2 x ) 2 sin ( 13 2 x ) cos ( 9 2 x )
3 π 4 , 7 π 4 3 π 4 , 7 π 4
0 , π 6 , 5 π 6 , π 0 , π 6 , 5 π 6 , π
3 π 2 3 π 2
No solution
0.2527 , 2.8889 , 4.7124 0.2527 , 2.8889 , 4.7124
1.3694 , 1.9106 , 4.3726 , 4.9137 1.3694 , 1.9106 , 4.3726 , 4.9137
Practice Test
sec ( θ ) sec ( θ )
− 1 2 cos θ + 3 2 sin θ − 1 2 cos θ + 3 2 sin θ
1 − cos ( 64 ∘ ) 2 1 − cos ( 64 ∘ ) 2
2 cos ( 3 x ) cos ( 5 x ) 2 cos ( 3 x ) cos ( 5 x )
4 sin ( 2 θ ) cos ( 6 θ ) 4 sin ( 2 θ ) cos ( 6 θ )
x = cos –1 ( 1 5 ) x = cos –1 ( 1 5 )
3 5 , – 4 5 , – 3 4 3 5 , – 4 5 , – 3 4
tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x ) tan 3 x – tan x sec 2 x = tan x ( tan 2 x – sec 2 x ) = tan x ( tan 2 x – ( 1 + tan 2 x ) ) = tan x ( tan 2 x – 1 – tan 2 x ) = – tan x = tan ( – x ) = tan ( – x )
sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x sin ( 2 x ) sin x – cos ( 2 x ) cos x = 2 sin x cos x sin x – 2 cos 2 x – 1 cos x = 2 cos x – 2 cos x + 1 cos x = 1 cos x = sec x = sec x
Amplitude: 1 4 1 4 , period: 1 60 1 60 , frequency: 60 Hz
Amplitude: 8, fast period: 1 500 1 500 , fast frequency: 500 Hz, slow period: 1 10 1 10 , slow frequency: 10 Hz
D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) D ( t ) = 20 ( 0.9086 ) t cos ( 4 π t ) , 31 second
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Unit 5 – Systems of Linear Equations and Inequalities
This unit begins by ensuring that students understand that solutions to equations are points that make the equation true, while solutions to systems make all equations (or inequalities) true. Graphical and substitution methods for solving systems are reviewed before the development of the Elimination Method. Modeling with systems of equations and inequalities is stressed. Finally, we develop the idea of using graphs to help solve equations.
Solutions to Systems and Solving by Graphing
LESSON/HOMEWORK
LECCIÓN/TAREA
LESSON VIDEO
EDITABLE LESSON
EDITABLE KEY
Solving Systems by Substitution
Properties of Systems and Their Solutions
The Method of Elimination
Modeling with Systems of Equations
Solving Equations Graphically
Solving Systems of Inequalities
Modeling with Systems of Inequalities
Unit Review
Unit #5 Review – Systems of Linear Equations and Inequalities
UNIT REVIEW
REPASO DE LA UNIDAD
EDITABLE REVIEW
Unit #5 Assessment Form A
EDITABLE ASSESSMENT
Unit #5 Assessment Form B
Unit #5 Assessment Form C
Unit #5 Assessment.Form D
Unit #5 Exit Tickets
Unit #5 Mid-Unit Quiz (Through Lesson #4) – Form A
Unit #5 Mid-Unit Quiz (Through Lesson #4) – Form B
Unit #5 Mid-Unit Quiz (Through Lesson #4) – Form C
U05.AO.01 – Solving Equations Graphically – Extra Practice (After Lesson #6)
EDITABLE RESOURCE
U05.AO.02 – Additional Modeling with Linear Systems
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Gina Wilson All Things Algebra Answer Key | Gina Wilson All things Algebra 2015
In the realm of mathematics education, finding reliable resources that support effective learning can be a challenging task. However, Gina Wilson All Things Algebra is a comprehensive platform that provides educators and students with valuable tools to enhance their mathematical knowledge. One of the key features of this platform is the availability of the answer key, which serves as a vital resource for learners seeking to validate their solutions and progress in their mathematical journey. In this article, we will delve into the benefits of Gina Wilson All Things Algebra and explore how the answer key can be accessed and utilized effectively.
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Unit 1: Equations & Inequalities Homework 3: Solving Equations page document! ** 2-3.96-23) 2.-3-9(5-2k) Your solution's ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on.
Introduction to Systems of Equations and Inequalities; 7.1 Systems of Linear Equations: Two Variables; 7.2 Systems of Linear Equations: Three Variables; 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables; 7.4 Partial Fractions; 7.5 Matrices and Matrix Operations; 7.6 Solving Systems with Gaussian Elimination; 7.7 Solving Systems with Inverses; 7.8 Solving Systems with Cramer's Rule
Find step-by-step solutions and answers to enVision Algebra 1 - 9780328931576, as well as thousands of textbooks so you can move forward with confidence. ... Section 1-3: Solving Equations with a Variable on Both Sides. Section 1-4: Literal Equations and Formulas. ... Section 8-1: Key Features of a Quadratic Function. Section 8-2: Quadratic ...
EXAMPLE 3: WRITE AND SOLVE A SYSTEM OF EQUATIONS. The sum of two angles is 180°. The measure of one angle is 34° greater than the measure of the other angle. Define the variables and write equations to model the situation. Let x and y be angles. ⎧ ⎪ x + y = 180 ⎨ ⎩⎪ x = y + 34. Find the measure of each angle.
2.1 Use a General Strategy to Solve Linear Equations; 2.2 Use a Problem Solving Strategy; 2.3 Solve a Formula for a Specific Variable; 2.4 Solve Mixture and Uniform Motion Applications; 2.5 Solve Linear Inequalities; 2.6 Solve Compound Inequalities; 2.7 Solve Absolute Value Inequalities
— Unit 1: Number Sense Homework 3: Multiplying & Dividing Integers 20 12 4(9) 6 . 5x-16 5(-6) 9 ... Gina Wilson All Things Algebra 2016 Unit 1 Answer Key WEBThe book addresses three important questions: 1. ... ** Solve each system of equations by graphing. Clearly identify your ... Answer Key Unit 1 Teaching Gifted Kids in Today's Classroom ...
Our resource for Big Ideas Math: Algebra 1 includes answers to chapter exercises, as well as detailed information to walk you through the process step by step. With Expert Solutions for thousands of practice problems, you can take the guesswork out of studying and move forward with confidence. Find step-by-step solutions and answers to Big ...
Unit 1 - Expressions, Equations and Functions 1.1: Order of Operations 1.2: Expressions, Equations, Inequalities 1.3: Functions as Rules and Tables 1.4: Functions as Graphs Unit 1 Review ...
Click the link below to take you there! to both sides. Finally, divide both sides by the coefficient of. Remember to always perform the same operation on both sides of the equation. 2\left ( {4 - y} \right) - 3\left ( {y + 3} \right) = - 11. Distributive Property of Multiplication over Addition.
Differences between Equations, Expressions and Equations. ≤ symbols Expressions DO NOT have =,>,<,≥, Equations ALWAYS have =. Inequalities have >,<,≥, ≤ symbols. EX: 2x + 1 is 2 TERMSIntepreting SolutionsIf you solve an equation and your solution is a vari. le equal t. a number, you have ONE solution. EX: x = -3If you salve an inequali.
In this course students will explore a variety of topics within algebra including linear, exponential, quadratic, and polynomial equations and functions. Students will achieve fluency in solving linear and quadratic equations as well as with manipulation of polynomials using addition, subtraction, multiplication, and factoring. Students will understand the key differences between linear and ...
Directions: Solve each equation. 3. 2(x + 4) = 2-X 8a + 20 - 5a ..3a- +20 2.0 7. 8 — 6m = 4m + 48 +1.4m +14m 10m -HOB 10m 8 + 16) = +30 Unit 2: Equations & Inequalities Homework 3: No Solution/lnfinite Solution SHOW ALL STEPS! 15 - = - -15 -gvJ= ZW -1595 W 4. 6. 8. 4(2r — 1) — — -2(3r + 16) 9+5 5v — 3p 11 10. 8x+ 11 = 2(4x-7) + 25 ...
provoke, and ignite change. Unit 5 - Systems of Equations & Inequalities (Updated … WEBThe bines 3) If a system of ... Name: Date: Unit 7: Exponential & Logarithmic Functions Homework 3: Intro to Logarithms Directions: Write each equation in exponential form. ... 2016 Unit 1 Answer Key WEBWithin the pages of "Gina Wilson All Things Algebra ...
ALGEBRA 1 CURRICULUM EQUATIONS & INEQUALITIES UNIT ONE: ANSWER KEY ©MANEUVERINC THE MIDDLE/ 2020 Unit: Equations and Inequalities Review Name Date EQUATIONS AND INEQUALITIES STUDY Solve each problem below. Be sure to ask questions if you need more help with a topic. I CAN SIMPLIFY ALGEBRAIC CXPPCSSIONS 1. Simplify an expression of the model. 2.
1-M Solving Inequalities 1-N Solving Inequalities Part 2 1-O Literal Equations Review Test All quizzes will be "pop quizzes" and there will be 2or 3 per unit. You MAY use this packet to complete the quizzes. This packet will be turned in on the day of the test for 100 points. Whenever you're absent, you can get these notes filled out from ...
1. The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures x, the second angle measures π 2 − x. Then sinx = cos(π 2 − x). The same holds for the other cofunction identities. The key is that the angles are complementary.
Select a Unit. Unit 1 Sequences; Unit 2 Linear and Exponential Functions; Unit 3 Features of Functions; Unit 4 Equations and Inequalities; Unit 5 Systems of Equations and Inequalities; Unit 6 Quadratic Functions; Unit 7 Structures of Quadratic Expressions; Unit 8 More Functions, More Features; Unit 9 Modeling Data
Unit 5 - Systems of Equations & Inequalities (Updated October 2016) copy. Name: Date: Unit 5: Systems of Equations & Inequalities Homework 1: Solving Systems by Graphing ** This is a 2-page document! ** Solve each system of equations by graphing. Clearly identify your solution. -16 — 6y = 30 9x + = 12 +4 v = —12 O Gina Wilson (All Things ...
Section 3.4 Solving Equations with Variables on Both Sides. Need a tutor? Click this link and get your first session free! Packet. 3.4_packet.pdf: File Size: 824 kb: File Type: pdf: Download File. Practice Solutions. 3.4practiceanswers1page.pdf: File Size: 137 kb: File Type: pdf: Download File. Corrective Assignment. 3.4_ca.pdf: File Size:
Unit 5 - Systems of Linear Equations and Inequalities. This unit begins by ensuring that students understand that solutions to equations are points that make the equation true, while solutions to systems make all equations (or inequalities) true. Graphical and substitution methods for solving systems are reviewed before the development of the ...
The answer key on Gina Wilson All Things Algebra offers various features that enhance the learning experience. Some notable features include: Detailed Solutions: The answer key provides comprehensive and detailed solutions to the exercises, enabling students to identify any errors and learn from them. Multiple Approaches: In many cases, the ...
HW #3 - Angle Relationships Answer Key - Free download as PDF File (.pdf), Text File (.txt) or read online for free. This 2-page document contains 17 geometry problems involving finding missing angle measures. The problems involve equations relating angle measures using variables like x and y. Several problems provide two angle measures and ask to find a third based on the information given ...