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2 years ago

The straight line ax+by=1passes through the points (1,4)and (3,1).Find the values of a and b?

Mathematics

2 answers:

Gelneren [198K]2 years ago

6 0

Answer:

a = $\frac{3}{11}$ , b = $\frac{2}{11}$

Step-by-step explanation:

Substitute (1, 4) and (3, 1) into the equation

a + 4b = 1 → (1)

3a + b = 1 → (2)

Multiplying (1) by - 3 and adding to (2) will eliminate a

- 3a - 12b = - 3 → (3)

Add (2) and (3) term by term to eliminate a

0 - 11b = - 2

- 11b = - 2 ( divide both sides by - 11 )

b = $\frac{2}{11}$

Substitute this value into (1) and solve for a

a + 4( $\frac{2}{11}$ ) = 1

a + $\frac{8}{11}$ = $\frac{11}{11}$ ( subtract $\frac{8}{11}$ from both sides )

a = $\frac{11}{11}$ - $\frac{8}{11}$ = $\frac{3}{11}$

Molodets [167]2 years ago

3 0

Answer:

a= 3/11

b = 2/11

Step-by-step explanation:

a(1) + b(4) = 1

a + 4b = 1

a(3) + b(1) = 1

3a + b = 1

b = 1-3a

a + 4(1-3a) = 1

a+ 4-12a = 1

-11a = -3

a = 3/11

b = 1-3(3/11)

b= 2/11

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I don’t really understand this type of stuff so please help.

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For 1 and 2, you plot both lines, and wherever they intersect is the solution to the system. Given the equation of a line, I think the easiest way to plot it is to find two points on the line, then draw a line through them. For example, if $y=5x-1$ , then when $x=0$ , you get $y=-1$ ; when $x=1$ , you get $y=4$ . So plot the points (0, -1) and (1, 4), then strike a line through.

1. Notice that dividing both sides of $2y=10x-2$ by 2 returns $y=5x-1$ , same as the first equation. So the system of equations reduces to one equation, which can have an infinite number of solutions. (This is because for any choice of $x$ or $y$ , you can always find a corresponding value for the other variable.)

2. See attached image. $3x-y=2$ is given by the purple line.

For 3-6, you have several options. The two simplest methods of solving them are by substitution or elimination.

3. Like with (1), notice that dividing both sides of the first equation by 2 gives $x+3y=9$ , so there will be an infinite number of solutions.

4. (by substitution) Since $y=-7x+3$ , we can replace $y$ in the second equation:

$-7x+3+7x=10\implies3=10$

but this is false, so there are no solutions to this system.

5. (by substitution) Since $x=2y+2$ , in the first equation we have

$-5(2y+2)+3y=-10y-10+3y=-7y-10=11\implies-7y=21\implies y=-3$

Then back in the second equation we find

$x=2(-3)+2=-6+2=-4$

So (-4, -3) is the only solution here.

6. (by substitution) Notice that the left hand sides of both equations are the same, so we end up with 7 = 12, but this is false, so no solution exists.