after state how many solutions there are and give a reason

please explain

thanks

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

Elimination Method:

- 2X + y = 4

3X + y =4

3X + 4Y = 5

-4(-2X- 4Y = 4 )

11 X = 5 -16 = -11

X = -11/ 11 = -1

Substitute X = -1 into 2nd equation

- 2( -1) + Y = 4

Y = 4 -2 =2

Substitution method

3X + 4Y = 5

-2X + Y = 4

Y = 2X +4 from 2nd equation:

3X + 4( 2X + 4) = 5

11X + 16 = 5

11X = -11 X = -1

Substitute in 2nd equation:

-2( -1) + Y = 4

Y = 2

To graph :

3X + 4Y = 5 3( 0) + 4Y = 5 Y intercept = 5/4 ( 0 , 5/4)

3 X + 4 (0 ) = 5 X intercept = 3/5 ( 3/5, 0)

Connect the 2 points together, and have the graph

- 2X + Y = 4 -2( 0) + Y = 4 ( 0 , 4 ) is Y intercept.

-2X + 0 = 4 (-2, 0 ) is X intercept

Connect 2 points and get the graph of the line

Observe that intersect point is :

( -1 ,2 )

3x+4y=5 => y = (5 - 3x)/4 = -3/4 x + 5/4

-2x+y=4 => y = (4 + 2x)/1 = 2 x + 4

The slopes are different, so the graphs will be two lines that intersect in one point, called the "solution" of the system of equations.

Solving by substitution:

y = -3/4 x + 5/4 = 2 x + 4

Multiply by 4:

-3 x + 5 = 8 x + 16

Add 3x-16 to both sides:

-11 = 11x

x = -1

y = 2 (-1) + 4 = 2

So the solution is (-1,2).

Solving by elimination:

3x+4y=5 => 3x+4y=5

-2x+y=4 => 8x-4y=-16

-2x+y=4 => 8x-4y=-16

11x = -11

x = -1

3x+4y=5 => 6x+8y=10

-2x+y=4 => -6x+3y=12

11y = 22

-2x+y=4 => -6x+3y=12

11y = 22

y = 2

So the solution, (-1,2), is the same as for substitution.

Solving using Cramer's Rule:

D = | 3 4 | = 3 - -8 = 11

| -2 1 |

D_x = | 5 4 | = 5-16 = -11

| 4 1 |

| 4 1 |

D_y = | 3 5 | = 12 - -10 = 22

| -2 4 |

| -2 4 |

x = D_x / D = -11/11 = -1

y = D_y / D = 22/11 = 2

So solution, as before, is (-1,2).

Solve by graphing:

One way is to convert each equation into Intercept Form:

ax + by = c => x/(c/a) + y/(c/b) = 1

where the intercepts are below their variable. Then graph the two intercepts and draw the line through them.

3x+4y=5 => x/(5/3)+y/(5/4)=1

-2x+y=4 => x/(4/-2)+y/(4/1)=1

-2x+y=4 => x/(4/-2)+y/(4/1)=1

See GeoGebra sketch here:

Hi Angelica;

3x+4y=5 and -2x+y=4

LINEAR COMBINATIONS

Both equations are linear combinations in that these are in the format of Ax+By, A and B are constants multiplying variables x and y.

3x+4y=5 is in Standard Formula...

Ax+By=C, neither A nor B equal zero and A is greater than zero.

-2x+y=4 is NOT in Standard Formula. A is less than zero. Let's fix that by multiplying both sides by -1...

(-1)(-2x+y)=(4)(-1)

2x-y=-4

SUBSTITUTION

3x+4y=5 and 2x-y=-4

Let's take either equation and isolate a variable. Obviously, it would be easiest to take the second equation and isolate y...

2x-y=-4

Let's subtract 2x from both sides...

2x-2x-y=-2x-4

-y=-2x-4

Let's multiply both sides by -1...

(-1)(-y)=(-1)(-2x-4)

y=2x+4

Let's take the first equation and substitute y with 2x+4...

3x+4y=5

3x+[(4)(2x+4)]=5

3x+8x+16=5

11x+16=5

Let's subtract 16 from both sides...

11x+16-16=5-16

11x=-11

Let's divide both sides by 11...

(11x)/11=-11/11

**x=-1**

Let's plug this into either equation to establish the value of y. I select the original second equation. It is easiest...

-2x+y=4

3x+4y=5 and -2x+y=4

LINEAR COMBINATIONS

Both equations are linear combinations in that these are in the format of Ax+By, A and B are constants multiplying variables x and y.

3x+4y=5 is in Standard Formula...

Ax+By=C, neither A nor B equal zero and A is greater than zero.

-2x+y=4 is NOT in Standard Formula. A is less than zero. Let's fix that by multiplying both sides by -1...

(-1)(-2x+y)=(4)(-1)

2x-y=-4

SUBSTITUTION

3x+4y=5 and 2x-y=-4

Let's take either equation and isolate a variable. Obviously, it would be easiest to take the second equation and isolate y...

2x-y=-4

Let's subtract 2x from both sides...

2x-2x-y=-2x-4

-y=-2x-4

Let's multiply both sides by -1...

(-1)(-y)=(-1)(-2x-4)

y=2x+4

Let's take the first equation and substitute y with 2x+4...

3x+4y=5

3x+[(4)(2x+4)]=5

3x+8x+16=5

11x+16=5

Let's subtract 16 from both sides...

11x+16-16=5-16

11x=-11

Let's divide both sides by 11...

(11x)/11=-11/11

Let's plug this into either equation to establish the value of y. I select the original second equation. It is easiest...

-2x+y=4

[(-2)(-1)]+y=4

2+y=4

Let's take both x and y results and plug these into the first equation for verification...

3x+4y=5

[(3)(-1)]+[(4)(2)]=5

-3+8=5

5=5

GRAPHING

I cannot do such here.

However,

3x+4y=5

2x-y=-4

The slope of each equation is -A/B...

3x+4y=5, -(3/4)=-3/4.

2x-y=-4, -(2/-1)=2

The y-intercept can be easily established as x=0...

3x+4y=5, 4y=5, y=5/4, y-intercept, (0,5/4)

2x-y=-4, -y=-4, y=4, y-intercept, (0,4)

When graphing, begin with the y-intercept. This is the point at which the line crosses the y-axis. For the first line, the line will increase 3 units as it runs to the left 4 units. For the second line, the line will increase 2 units as it runs to the right 1 unit. The two lines will insect at (-1,2).

ELIMINATION

This is another method you do not mention.

3x+4y=5 and -2x+y=4

To do this, either variable must have the same coefficient. Currently, x has the coefficients of 3 and -2, whereas y has the coefficient of 4 and 1.

Let's take the second equation.

-2x+y=4

Let's multiply both sides by 4.

On second thought, let's multiply both sides by -4 such that we convert this into Standard Formula...

(-4)(-2x+y)=(4)(-4)

8x-4y=-16

Let's add the two equations together and eliminate...

8x-4y=-16

+(3x+4y=5)

11x=-11

x=-1

SUBSTITUTION, GRAPHING AND ELIMINATION ARE ALL TECHNIQUES WHICH CAN BE USED TO SOLVE THIS.

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