How do you solve $y = 7 x - 1$ $y=7x-1$ and $y = - x + 14$ $y=-x+14$ using substitution?

Question

How do you solve $y = 7 x - 1$ $y=7x-1$ and $y = - x + 14$ $y=-x+14$ using substitution?

2 Answers

Impact of this question

6218 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-26T13:00:54

The solution is $(\frac{15}{8}, \frac{97}{8})$ $(15/8,97/8)$ or $(1.875, 12.125)$ $(1.875,12.125)$ .

Explanation:

Solve the system:

$Equation 1:$ $"Equation 1:"$ $y = 7 x - 1$ $y=7x-1$

$Equation 2:$ $"Equation 2:"$ $y = - x + 14$ $y=-x+14$

Both linear equations are in slope-intercept form. The solution to the system is the point that both lines have in common, the point of intersection. Substitution will be used to solve the system.

Both equations are set equal to $y$ $y$ . Substitute $7 x - 1$ $7x-1$ from Equation 1 for $y$ $y$ in Equation 2 and solve for $x$ $x$ .

$y = - x + 14$ $y=-x+14$

$7 x - 1 = - x + 14$ $7x-1=-x+14$

Add $x$ $x$ to both sides.

$x + 7 x - 1 = 14$ $x+7x-1=14$

Simplify.

$8 x - 1 = 14$ $8x-1=14$

Add $1$ $1$ to both sides.

$8 x = 15$ $8x=15$

Divide both sides by $8$ $8$ .

$x = \frac{15}{8}$ $x=15/8$ or $1.875$ $1.875$

Substitute $\frac{15}{8}$ $15/8$ for $x$ $x$ in Equation 1 and solve for $y$ $y$ .

$y = 7 x - 1$ $y=7x-1$

$y = 7 (\frac{15}{8}) - 1$ $y=7(15/8)-1$

$y = \frac{105}{8} - 1$ $y=105/8-1$

Multiply $1$ $1$ by $\frac{8}{8}$ $8/8$ to get a denominator of $8$ $8$ .

$y = \frac{105}{8} - 1 \times \frac{8}{8}$ $y=105/8-1xx8/8$

$y = 105 - \frac{8}{8}$ $y=105-8/8$

Combine the numerators.

$y = \frac{105 - 8}{8}$ $y=(105-8)/8$

$y = \frac{97}{8}$ $y=97/8$ or $12.125$ $12.125$

The solution is $(\frac{15}{8}, \frac{97}{8})$ $(15/8,97/8)$ or $(1.875, 12.125)$ $(1.875,12.125)$ .

graph{(y-7x+1)(y+x-14)=0 [-9.455, 10.545, 5.48, 15.48]}

Answer 2 · 2018-04-26T13:02:13

$(x, y) : (\frac{15}{8}, \frac{97}{8})$ $(x, y): (15/8, 97/8)$

Explanation:

$e q_{1} : y = 7 x - 1$ $eq_1 : y=7x-1$
$e q_{2} : y = - x + 14$ $eq_2 : y=-x+14$

Use $7 x - 1$ $7x - 1$ for y in $e q_{2}$ $eq_2$ :
$7 x - 1 = - x + 14$ $7x - 1 = -x + 14$
$8 x = 15$ $8x = 15$
$x = \frac{15}{8}$ $x = 15/8$

$y = 7 x - 1$ $y=7x-1$
$y = 7 \times \frac{15}{8} - \frac{8}{8}$ $y = 7xx15/8-8/8$
$y = \frac{97}{8}$ $y = 97/8$

Science

Math

Humanities

... and beyond

How do you solve $y = 7 x - 1$ $y=7x-1$ and $y = - x + 14$ $y=-x+14$ using substitution?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve y=7x−1y=7x-1 and y=−x+14y=-x+14 using substitution?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you solve $y = 7 x - 1$ $y=7x-1$ and $y = - x + 14$ $y=-x+14$ using substitution?