Login
Premium
Home Question Bank Math Algebra
Question
upstudy study bank question image url

What is the solution to the system of equations? \( y=\frac{1}{8} x-1 \) \( -5 x+4 y=-13 \)

Ask by Frazier Murphy. in the United States
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The solution to the system of equations is \( x = 2 \) and \( y = -0.75 \).

Solution

Solve the system of equations \( y=\frac{1}{8}x-1;-5x+4y=-13 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=\frac{1}{8}x-1\\-5x+4y=-13\end{array}\right.\) - step1: Substitute the value of \(y:\) \(-5x+4\left(\frac{1}{8}x-1\right)=-13\) - step2: Simplify: \(-\frac{9}{2}x-4=-13\) - step3: Move the constant to the right side: \(-\frac{9}{2}x=-13+4\) - step4: Add the numbers: \(-\frac{9}{2}x=-9\) - step5: Change the signs: \(\frac{9}{2}x=9\) - step6: Multiply by the reciprocal: \(\frac{9}{2}x\times \frac{2}{9}=9\times \frac{2}{9}\) - step7: Multiply: \(x=2\) - step8: Substitute the value of \(x:\) \(y=\frac{1}{8}\times 2-1\) - step9: Simplify: \(y=-\frac{3}{4}\) - step10: Calculate: \(\left\{ \begin{array}{l}x=2\\y=-\frac{3}{4}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=-\frac{3}{4}\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(2,-\frac{3}{4}\right)\) The solution to the system of equations is \( (x,y) = (2,-\frac{3}{4}) \) or \( (x,y) = (2,-0.75) \).

Reviewed and approved by the UpStudy tutoring team

error msg
Explain
Simplify this solution

The Deep Dive

To solve the system of equations, start by substituting the expression for \( y \) from the first equation into the second equation. This gives us: \[ -5x + 4\left(\frac{1}{8}x - 1\right) = -13 \] Next, distribute the 4: \[ -5x + \frac{4}{8}x - 4 = -13 \] which simplifies to: \[ -5x + \frac{1}{2}x - 4 = -13 \] Now, combine the \( x \) terms: \[ -\frac{10}{2}x + \frac{1}{2}x = -13 + 4 \] This simplifies to: \[ -\frac{9}{2}x = -9 \] Now, to isolate \( x \), multiply both sides by \(-\frac{2}{9}\): \[ x = 2 \] Now that you have \( x \), substitute it back into the first equation to find \( y \): \[ y = \frac{1}{8}(2) - 1 = \frac{1}{4} - 1 = -\frac{3}{4} \] So the solution to the system of equations is: \[ (x, y) = (2, -\frac{3}{4}) \]

Related Questions

7 Expand and simplify \( 5 x(3 x+4)(2 x-1) \)
Algebra United Arab Emirates Jan 22, 2025
2) \( a^{2}-11 a-12 \)
Algebra South Africa Jan 22, 2025
5. \( \begin{aligned} j-9 \pi= & 12 \pi \\ +9 \pi & +9 \pi\end{aligned} \)
Algebra United States Jan 22, 2025
An electrician charges an initial fee of \( \$ 50 \) and \( \$ 190 \) after 4 hours of worl a. Write a linear model that represents the total cost as a function of the \[ y=35 x+50 \] b. How much does the electrician charge per hour? \[ 4 x+x=140+5= \]
Algebra United States Jan 22, 2025
4. \( \frac{2 x-1}{3}-\frac{4 x-8}{6} \) 6. \( \frac{3}{x-1}+\frac{1}{1-x} \) 8. \( \frac{x}{x+y}+\frac{y}{x-y} \) 10. \( \frac{x+3}{x-3}-\frac{x-3}{x+3} \) 12. \( \frac{5}{3(a-b)}+\frac{3}{2(b-a)} \) 14. \( \frac{x}{x^{2}-y^{2}}+\frac{y}{y^{2}-x^{2}} \) 16. \( \frac{4 x}{(x+y)^{2}}-\frac{4}{x+y} \) 18. \( \frac{3 a}{9 a^{2}-4 b^{2}}-\frac{1}{3 a+2 b} \) 20. \( \frac{2 y}{(x-2 y)^{2}}+\frac{1}{x-2 y} \)
Algebra United States Jan 22, 2025

Latest Algebra Questions

2. Find the products. \( \begin{array}{ll}\text { a) }(x+1)(x+3) & \text { b) }(a+3)(a+4) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { e) }(3 x y+2 z)(2 x y-3 z) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { k) }(2 a-b)(2 a+b) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \text {-n) }(x+3)^{2}\end{array} \)
Algebra South Africa Jan 22, 2025
5. \( \begin{aligned} j-9 \pi= & 12 \pi \\ +9 \pi & +9 \pi\end{aligned} \)
Algebra United States Jan 22, 2025
7 Expand and simplify \( 5 x(3 x+4)(2 x-1) \)
Algebra United Arab Emirates Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { (b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { 1) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b)\end{array} \)
Algebra South Africa Jan 22, 2025
1. Find the products. \( \begin{array}{ll}\text { a) }(x+2)(x+3) & \text { \b) }(p-3)(p-4) \\ \text { c) }(a-2)(a+3) & \text { d) }(2 x-3)(3 x+1) \\ \text { e) }(2 y+5)(3 y-2) & \text { f) }(3 a-7)(2 a+1) \\ \text { g) }(2 a-3)(3 a-2) & \text { (h) }(4 y+1)(7 y-2) \\ \text { i) }(x-3)(x-2) & \text { j) }(4 a-2)(3 a+5) \\ \text { k) }(2 x-3 y)(2 x+4 y) & \text { \l) }(2 a-7 b)(3 a-2 b) \\ \text { m) }(7 x+3 y)(2 x-4 y) & \text { n) }(3 a-2 b)(2 a+11 b) \\ \text { 2. Find the products. } & \text { b) }(a+3)(a+4) \\ \text { a) }(x+1)(x+3) & \text { (d) }\left(a^{2}-2\right)\left(a^{2}+2\right) \\ \text { c) }\left(x^{2}-1\right)\left(2 x^{2}+3\right) & \text { f) }(p r+2 t)(3 p r-t) \\ \text { (e) }(3 x y+2 z)(2 x y-3 z) & \text { h) }\left(2 a^{2}+1\right)\left(3 a^{2}+1\right) \\ \text { g) }\left(x^{2}-4\right)\left(4 x^{2}-3\right) & \text { j) }\left(2 x+\frac{1}{3}\right)\left(3 x-\frac{1}{2}\right) \\ \text { i) }\left(x+\frac{1}{2}\right)\left(x+\frac{1}{4}\right) & \text { l) }(3 x+2 y)(3 x-2 y) \\ \text { k) }(2 a-b)(2 a+b) & \text { n) }(x+3)^{2} \\ \text { m) }\left(\frac{1}{2} x+\frac{1}{3} y\right)\left(\frac{1}{3} x-\frac{1}{2} y\right) & \end{array} \)
Algebra South Africa Jan 22, 2025
Prev Next
Try Premium now!
Try Premium and ask Thoth AI unlimited math questions now!
Maybe later Go Premium
Study can be a real struggle
Why not UpStudy it?
Select your plan below
Premium

You can enjoy

Start now
  • Step-by-step explanations
  • 24/7 expert live tutors
  • Unlimited number of questions
  • No interruptions
  • Full access to Answer and Solution
  • Full Access to PDF Chat, UpStudy Chat, Browsing Chat
Basic

Totally free but limited

  • Limited Solution
Welcome to UpStudy!
Please sign in to continue the Thoth AI Chat journey
Continue with Email
Or continue with
By clicking “Sign in”, you agree to our Terms of Use & Privacy Policy