9th grade algebra 1 math help?

HELP! i can't figure any of these out! you can answer one, a few, or all of them. & tell me how you got the answer

thanks! xoxo

A farmer kept track of his cows and chickens by counting both the legs and heads. If he counted 78 legs and 35 heads, how many cows and chickens did he own?

There are only bicycles and tricycles in kyle's backyard. he counted a total of 30 seats and 70 wheels in the backyard. how many tricycles are in his backyard?

a fraction is equivalent to 3/5. its denominator is 60 more than its numerator. what is the numerator of this fraction?

if, on average, a hen and a half lays an egg in a day in a half, how many effs will 6 hens lay in 8 days?

13 plums weigh as much as 2 apples and 1 pear. 4 plums and 1 apple have the same weight as 1 pear. how many plums have the weight of 1 pear?

Comments

  • You know a cow has 4 legs and a chicken has 2.

    Let W = the number of cows and K = the number of chickens.

    Then you know that the number of cows times 4 legs plus the number of chickens times 2 legs is 78 legs altogether, or in algebraic terms:

    4W + 2K = 78

    You also know the number of cows plus the number of chickens is 35:

    W + K = 35

    Take the second equation and isolate one of the variables (arrange the equation so that one letter is on one side of the equals sign):

    W + K = 35

    Subtract K from both sides:

    W + K - K = 35 - K

    W + 0 = 35 - K

    W = 35 - K

    Now, since W is the same as 35 - K, you take 35 - K and substitute it for the W in the first equation:

    4W + 2K = 78

    4(35 - K) + 2K = 78

    Distribute the 4 and solve for K:

    140 - 4K + 2K = 78

    140 - 2K = 78

    Add 2K to each side and subtract 78 from each side:

    140 - 78 = 2K

    62 = 2K

    Divide both sides by 2:

    31 = K

    Now substitute 31 for K in the second equation:

    W + K = 35

    W + 31 = 35

    Subtract 31 from each side:

    W = 4

    So 4 cows and 31 chickens.

    The second question is just like the first. I will leave that for you to solve.

  • A farmer kept track of his cows and chickens by counting both the legs and heads. If he counted 78 legs and 35 heads, how many cows and chickens did he own?

    Given:

    Each cow has 4 legs; each chicken has 2 legs; that is, they all have their legs intact.

    Each cow has 1 head; each chicken has 1 head

    Let:

    x = number of cows

    y = number of chickens

    Equation for legs:

    4x + 2y = 78

    Equation for heads:

    x + y = 35

    OR

    x = 35 - y

    Substitute this to the first equation, then solve for y:

    4(35 - y) + 2y = 78

    140 - 4y + 2y = 78

    -4y + 2y = 78 - 140

    -2y = -162

    y = -62/-2

    y = 31

    Substitute to the second equation then solve for x:

    x = 35 - y

    x = 35 - 31

    x = 4

    Check using the first equation:

    4x + 2y = 78

    4(4) + 2(31) = 78

    16 + 62 = 78

    78 = 78 Yes!

    Answer:

    The farmer owns 4 cows and 31 chickens.

    There are only bicycles and tricycles in kyle's backyard. he counted a total of 30 seats and 70 wheels in the backyard. how many tricycles are in his backyard?

    Given:

    Each bicycle has 1 seat and 2 wheels.

    Each tricycle has 1 seat and 3wheels.

    Let:

    B = number of bicycles

    T = number of tricycles

    Seats:

    B + T = 30

    OR

    B = 30 - T

    Wheels:

    2B + 3T = 70

    Substitute the first equation to the second, then solve for T:

    2B + 3T = 70

    2(30 - T) + 3T = 70

    60 - 2T + 3T = 70

    -2T + 3T = 70 - 60

    T = 10

    Answer:

    There are 10 tricycles in Kyle's backyard. There are 20 bicycles.

    The checking is left to you.

    a fraction is equivalent to 3/5. its denominator is 60 more than its numerator. what is the numerator of this fraction?

    Let:

    N = the numerator

    D = The denominator

    Eqn1:

    ...N..........3

    -------- = --------

    ...D..........5

    Eqn2:

    D = N + 60

    Substitute the second equation to the first, then cross-multiply, then solve for N:

    ...N..........3

    -------- = --------

    ...D..........5

    .......N...............3

    ----------------- = --------

    ...(N + 60)..........5

    5N = 3(N + 60)

    5N = 3N + 180

    5N - 3N = 180

    2N = 180

    N = 180/2

    N = 90

    D = N + 60

    D = 90 + 60

    D = 150

    ...N..........90...........3

    -------- = ----------- = -------

    ...D..........150.........5

    Answer:

    The numerator of the fraction is 90; it's denominator is 150.

    if, on average, a hen and a half lays an egg in a day in a half, how many effs will 6 hens lay in 8 days?

    Let:

    E = number of eggs

    (1 egg / 1.5 hens) / (1.5 days) = (E / 6 hens)( / (8 days)

    E = (1 / 1.5) / (1.5) * (6)(8)

    E = 21 eggs <=== Answer

    13 plums weigh as much as 2 apples and 1 pear. 4 plums and 1 apple have the same weight as 1 pear. how many plums have the weight of 1 pear?

    Let:

    P = weight of a pear

    a = weight of an apple

    p = weight of a plum

    x = number of plums equivalent to 1 pear

    Equations:

    13p = 2A + P...........13 plums weigh as much as 2 apples and 1 pear.

    4p + A = P..............4 plums and 1 apple have the same weight as 1 pear.

    Required:

    P = xp

    From the second equation:

    A = P - 4p

    Substitute to the first equation:

    13p = 2A + P

    13p = 2(P - 4p) + P

    13p = 2P - 8p + P

    2P + P = 13p + 8p

    3P = 21p

    P = 7p

    Answer:

    7 plums have the weight of 1 pear.

    To check:

    Solve for A using the first equation:

    13p = 2A + P

    2A = 13p - P

    2A = 13p - 7p

    2A = 6p

    A = 3p

    Solve for A using the second equation:

    A = P - 4p

    A = 7p - 4p

    A = 3p Yes!

  • 1. Assume that cows = x and chickens = y.

    Both cows and chickens only have one head, so:

    x + y = 35

    Since cows have 4 legs and chickens have 2:

    4x + 2y = 78

    y= 35 - x

    4x + 2(35 - x) = 78

    4x + 70 - 2x = 78

    2x = 8

    x = 4

    x = the number of cows, so if there are 4 cows, there are 31 chickens.

    2. Assume that x = bicycles, and y = tricycles.

    x + y = 30 and 2x + 3y = 70

    2(x + y = 30)

    2x + 2y = 60 and 2x + 3y = 70

    (subtract the second equation from the first)

    -y = -10 so y = 10

    There are 10 tricycles, so there are 20 bicycles.

    3. 3/5 = x/(x + 60)

    3/5(x + 60) = x

    3/5x + 36 = x

    36 = 2/5x

    90 = x

    4. If 1.5 hens lays an egg in 1.5 days. 6 hens would lay 4 eggs in 1.5 days. So, they would lay 5 and 1/3 eggs four times in 8 days, bringing your total to 21 and 1/3 eggs in 8 days.

    You'll have to do the last one yourself. It's just a simple substitution problem, but I'm tired so good luck. Hope this helped.

  • very nicely you need to locate 2 numbers that provide you the coefficient of the 1st and final time era while more suitable and while added jointly provide you the middle words coefficient.... there is a few thing suggested as the container technique write the climate of the 1st coefficient interior the 1st Q it incredibly is a million a million(x)^2 is your first time era..... a million -a million those are your components a million and a million or -a million and -a million a million -a million try this for the final time era 6 a million 2 those are your components a million and six and 2 and 3 adverse 6 3 components are includedtoo merely theres too many to jot down.. now u ought to locate the pair of components from the 1st time era and the 2d time era that as quickly as more suitable jointly... and added gives you you the middle time era... subsequently 5x so they ought to multiply and upload to 5 a million 3 it is trial and blunder a million 2 make a *container* which includes your words and circulate multiply so u get 3 x1 and 2 x1 once you multiply them out you get 3 and 2 wich upload to 5 your center time era so which you recognize your nicely suited.... then you incredibly merely place crackets around your components (a million 3) (a million 2) those at the instant are your 2 components.... the sign interior corresponds to the sign of the your components. subsequently there the two valuable so.. (x+3)(x+2)=0 set each to 0 x+3 = 0 or x = -3 x+2 = o x = -2

  • a fraction is equivalent to 3/5. its denominator is 60 more than its numerator. what is the numerator of this fraction?

    x/(x + 60) = 3/5

    5x = 3(x + 60) {cross-multiply}

    5x = 3x + 180

    2x = 180

    x = 90

    check

    90/150 = 9/15 = 3/5

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